8L2C)pJJJJ IH(ȱH:=IH[H`@HcH  $ +   I/H`JLNGȄBȄF aK  haaFF  mJm# KKJ UJ )J ۈ) ;J3ȱJFȱJGJKaȄM  aaNNJFLGJL L&FORCEVELOCITY%VELOCITY LENGTHVELOCITY SPEED'VELOCITY VELOCITYCVELOCITY VOCABCVELOCITY REVIEW DACCELERATIONEACCEL ACCEL7yACCEL CONSTANT!ACCEL NEWT1ݛACCEL NEWT2ACCEL VOCAB!ACCEL REVIEW@FRICTIONSCICTRL.TXT& HELP.OVRf*GFUTIL.OVRj 8f*MENU.OVRVf*MINI.HLP'f*#% -*PDISK1.CCP0?-1DPDISK1.CRV-1DPDISK1.SFG" 0&PDISK1&H#  'SCI.FNT% ~HIRES.TBL# EPDISK1' PDISK1.CLC-1D*PDISK1.FIG +S77*PDISK1.CVC  PDISK1.CVL *PDISK1.LBL4Q% -*PDISK1.LLC7꽌ɪ꽌ɖ (I%ɠ@*<%<(=Ю@AKV<Y<<Y<&，YЬV0&^*^*&+`L ::RF? &PRODOS `DaElH$?EGvѶK+`L HHLy XP LM ŠϠĠӠS)*+,+`F)) (*=GJFjJJA QE'+ '== `@ STSP8QSS8 m P o R(8R˾FRICTION ADD\FRICTION FRICTIONFRICTION RESOLVEFRICTION INCLINE*6FRICTION VOCABP6FRICTION REVIEWq6EXPERIMENTz=EXPBAL=ONE=ONE DONE>VOCAB:>REVIEWU>MORE00`$00{JE=Z6[ot6kcxuy4s[8,i  !"#$%&'()*+,-./ ?`pp8X8p}pn`y?  x Xxx0 0@x`x| @@@@@ @@@'AOXXXYXLXY YXLXY@`@Y3Y3Y W  D|_  `~(@@@@@0@p@@@@@@@@@x@@@@@@@@@?llg@@@p@0@@@@@x@@@@??11>?>55?755?7551111?55 ~| 0 |~ LLG?>11111?6550 |dnvfXX0 0@x`x|  ``0@`p  )P@ 9  . ( 5 2P@@ / 8 ( +`` <P@@ $ C (  HP@@@  Q (  0xL@p@Lx XP @  h h h Cw`x`@``~@``@@``@`UP a *( ` U@ ^ *( ]  T@ Z   ( X  < TP@ U > 8 ( S  fP@ O  L ( M00 pyqP@ I  ( F 0 0!P@ A <0& ( >? 2| |@@Lx LLL|@@ xL@p@Lx xL@`0| `p `p )pp p q pp pq p q p p c c c c c c  ?          || ^ s3 UU c *( b  *** (* 0``@3@``0 UUU 000```@ @```000 TPP@@6CJ   * >1111%H$ `0 0` 00p00pp000`0p0 0ppp 20??0 ???>3300?00033??]<`*** xcxUUU pgp*** @P_kUUUx~{}{owNxpX | `***pp_?_@***@i@ UUUxgx ***8`a`8 UUU8`[`8***p@Q@pUUU 0`@E@`0*(000 '66yA`; ((* )  ;UUUEEEeee555 000```@@#@@@ @@`00 # 00`@@ (*( @`0 ; 0`@ UUU@p K p@ ***`8 U 8`UUU`<`8 U 8``<]<` xcx pgp @P_kx x~{}{owNxh ???g@ @i@ xgx 8`a`8 8`[`8p@Q@p0`@E@`0 0``@3@``0 ' 000```@@``` p`@@`p 999. AAAAAA@`pp03 ?mL-  3AAAAA. AAAAAA10pp`@ - Lm? 1 AAAAA.@@```000 000```@@."@@`00 # 00`@@"@`0 ; 0`@@p K p@@{?owNx `x?||nvn<\{8>`" `pp8X8pp`?}ny?999@@@@ p88p   !"#$%&'()+,-./;<=>|@@`}`~`}@w@;g| p|~?~?w{w^?n=p @pxx\l\x~8wp|@ ~@``p~0p~`{`]@s~ x~@{?@}?|W *"`<** U pUU ~*@@x*@*@@@U`<UU*p ** xU@x UU x*`<**@@|UpUU 8`*@x** p@U `<UU  0` * p**  ``@@@U8UU?>55?755?755111111?55 ~| 0 |~ |l<<`| 0 |~|p@ *xU  00```U```pxp ** "UU > 8 "**f "UUL "**pyq "UU "**00 "U<0 D  W@```@``@`x````@@``@@``````x`000*00``GLL 8`UU p@  00``@ *@@* p 0`@UU `<x@**@@``0  pU@`0 Ux@`0* `8 * @```00 U ~CU`8 * @@ `0` @@sc Fc F c Bz*Bb2+@@`0U p U @`000 * ?a*@pU 8`@ U@`@` |*000 `@``@`x````@@``@@``````x````@@``@|````  ?ffc   ~| 0 |~ |l<<`| 0 |~ `0@`p @@ `0`@ 0`@Ux@*0p!pU@`0* @```00 U`8 *xU<>fff<*@ #UCOCCCN#*y#U3#* 8"**/*~@`pyp3~yp?~pI3pOA3pg`~?`@@~x@|?p@?x~@|g?@Oyg?@yg8@|'N9@?~N9@g8xg|9~~~x@~_W~~~|/kz~yyIA?P@P33? |@px|a?~|L?~|O?fA|g?rL|s?pL|y?fAx@?fOxapp@~p|?p`||p|?/~x_Wu}ss333 ( ?gg~?!@px|xxpp@@|sqsa~p_W5?#kz~~~frpff @P?ALLAOa (  P@P r @`p``@@p~3?Og`p|~?|O?|O|O9p|Os|s|O9p>xOys~?8|?|p@|`Y@px|xxpp@|sqsa~p?~~frpffALLAOam @`@@xgLxgdxg`xgLpCLpC``?|`~x`|@`s`s`3`g`g`3@~3~@|p?x~x`~p@~`x?|~y~x~ys`~y9f~y8f~ys`|psg|pxx?`x?@~x?`pxg|xc|p~@`pOypGypOpOI3pOA3pO`?`@@~x@|?p@?x~@?~g?@~g?@?~g8@?~'N9@?~N9@?~g8|g|9~*| ] *}```+U 3333333|@px|s?~|q?~|s?fA|s?rL|s?pL|s?fAxa?fOxapp@~p|? p8X@@@`@@@@p`0773 0LLLOL ,,33x00~*U* } _ |*U:`@U 0`@* 0`@U @@``a36@@`j00  @@@```00 U*U`*U*?U``*U ,,,,,,,&3x |x xs  p,,L |8,,x8,,x x xx,,LGg@@@@@A;oof`? ?`?@@@@3 3 3p~@p*`<U @x* |U   *  U  *  0`@U  *xU `8*p@`0x`0 U```03@p*<f< `8U `8*@`@`U``` AO"*xxq"U" ``p```@'030003)~f)LLL )& (&$, x@@x~fppp`XX`xx Y 3p^xLx @* U `<*<f<@`8U @@@L @`8 *`0 U @p*  `0U `03c|*pU  8`@*080x  00`@ U@@@L xf"UU/UL"**/*pyq"UU/U"**/*00"/<0t X&.```~>cc>||;oof>cc>33 x p0@0p x    ` @`@`LLLOL LLFCAx`~p@~`x?|~p>f~gs`~s9f~y8f~|s`<`sg|pxx?`x?@~x?`pxC|x|xLxOdxg`xsLpLpC``?|`~x`|@`~s`g|s`|3`?~g`g`O3@|3~@|p?x` 00``@@ 0``@@0pp`@@`pp0$ 6+'03#`f@ @  0 3`f@ L 03`fp@ L   n33``@@"   (((( (  @PTUUUE@EDDUP@ ( ***((* * *"(""("*(  UWW U}}UTQUPWDUQDQUU   * *((""("*    U U               L X0@ ~pB )~ 0``@@@G|a`4 p p x@ <`>`p0aOx@|$1 @`0p  x p x cgo{scc accacca @`pxxp`@   @`px<~~`x~:@p|?7`x~3@p|?0`x~,@p|?" (((*jx~?/! $T Uuu}_WUT"@`jz::>>.//+/?~zj*"``@``px8UUUUUU?}u#pxx\^OG?|p@!  ~x`% ?|p@@@P1  ( 8 PPP@PP8@@@@@@*..(((;@ACGO^=9E YY 2"@0 LL X0@!~`0 )~!Fx@ACGO^=9E YYYYYL hx E@@P @ (  9@@P  4 (  -@@P ( (   !@@P%  ( , 0 LL X0@<~[!~@ )~!0``@@@@a`4,....+0``0-YY@``N<~[Cxx C N N N N N N N N N I@@@@@@I??YYYYYNpp N` &  &,`D 00`00 pp`0,  ,0`|OAS| 0`@@B@h0 p@@pv 7@"@p|~??xOsOsO9pyOs|s?~O9ppOys~?8|?|p@|`i @0 LL X0@~p )~0``@@@@a`48`@@`8a @`@>n33``@@  00``@@ 3  0v`@@0pp`@@`pp0!L ! | 3 p f ` L@< x3pf`L@<x3pf`L@<x`Xygp2Z+''030##`f`@ @  @ 0 30`f`@ L @  03 0`f$`pp@ L !@Llx@O|`R?~pTx@V|` Y?~p [x@]|``?~pbx@d|gq6? <||. || x@+|`)?~p'x@$@@``pp88<<..//~p?|``@@@ ||^^W UW|a~p<?~p@@``pp88x@9|`7/?~p51x@23|`06?~p.8x@+:|`)=?~p'?x@$A|`"D?~p Fx@H|`K?~pM`x~&@p|?* pp`x~- |oc```.@@@@@@@@@@@@@@Y&qU`x@N|`L~pJ @@``pp88~p?x@G||N ?xC|`E|`|a~pC ?~p~p?x@@ x@?xC|`>|`_UUUUUUUUUUU  `@@`  ~>  ) " \fff| & + & R R R A%xx%| |  xL >  @`q`@  pp 0 0pUUUUU p00p000 ``00 80pP0888<44b88?888? 80`@@``p``` W U U`PPXW "UW    * 1*  U 1U  * *zz +*GG  U U +U  * * +* u U}eee }UUU UUU}eee } Uu    4@p||~~ ?@@`PPp .@@` P8W/_? Ow{}~x`x||~~?n** &p\,V*U+U*U*U*U*U*U*? ?u @``` @jUj5*5k*U*U:  @`px<~~=!g R R R L`P`@ G`p|~}z:j*+~ C@x~   <@@`ppxx ? )@ x 03|0`f|`pp@ L |@Ll<<~>n33``@@ |@@`0  00``@@ 3 | @@`0  0v`@@0pp`@app0@@`0 x93p;f`<L@<>x`Xygp?>qY+&'0300"#`ff`@ @  @ @ 0 30 0`f~P `@ L s  <<~>n33``@@ @@`0  00``@@ 3 @@`0  0v`@@0pp`@@`pp0@@`0 !L 7 "! x+ 3 p-f `.L@<0x23p4f`5L@<7xC~p @@``00 m?~j*+?\QOwpp0 p10p000?u Uuu]W[{p>|`pp000zj*+/zj*/?xC~p  }uUW}uUW_|ax@  ?~z***j{~z:|`ADDV3IADDV4TADDV5_ADDV6jADDV7vADDV7BADDV8ADDV9ADDV10ADDV10BADDV11ADDV12ADDV12BADDV12CADDV12DFRICTN1FRICTN2 RESOLV1RESOLV3&RESOLVED4/ ball1b2 VECTORS1.VECTORS2<VECTORS3JVECTORS4XVECTORS5dACCEL1pACCEL2|ACCEL3HARMONY1HARMONY2HARMONY3CRAZYPCRAZYVCRAZYABIGBALLBIGBALL1BIGBALL2BIGBALL3BIGBALL4L\RESOLV1\ \L\RESOLV3\ \L\RESOLVED\ \L\INCLTEST\ \L\FINCLIN1\ \L\V5OLD\ 35 \L\BIGBALL4\ \L\BIGBALL5\ \L\ADDV1\ \L\ADDV2\ \L\ADDV3\ \L\ADDV4\ \L\ADDV5\ \L\ADDV6\ \L\ADDV7\ \L\ADDV7B\ \L\ADDV8\ \L\ADDV9\ \L\ADDV10\ \L\ADDV10B\ \L\ADDV11\ \L\ADDV12\ \L\ADDV12B\ \L\ADDV12C\ \L\ADDV12D\ \L\FRICTN1\ \L\FRICTN2\ \\L\ball1\ \L\b2\ \L\VECTORS1\ \L\VECTORS2\ \L\VECTORS3\ \L\VECTORS4\ \L\VECTORS5\ \L\ACCEL1\ \L\ACCEL2\ \L\ACCEL3\ \L\HARMONY1\ \L\HARMONY2\ \L\HARMONY3\ \L\CRAZYP\ \L\CRAZYV\ \L\CRAZYA\ \L\BIGBALL\ \L\BIGBALL1\ \L\BIGBALL2\ \L\BIGBALL3\ts learn about the laws of physics.\ \DCW\ \SWT\\\6\\ \DTNNW\ When an object moves, it moves from one place to another, over some period of time.\ \SA\ \DFA\BALL1\8\8\ \SAS\50\1\1\-1\ \SAL\26\10\0\0\0\ \DA\ \DT\\ \DTN\ When an object moEW\ \BL\FORCE\ \L\VELOCITY LENGTH\ \SCH\2\2\ \DTCNW\Length and Time\ \SCH\1\1\ \DTN\ The central theme of all physics is the movement of matter. Without movement, there would be no study of physics.\ \DTN\ By studying motion, physicis Time\ \SMO\Speed\ \SMO\Velocity and Vectors\ \SMO\Vocabulary Review\ \SMO\Review Questions\ \SMO\Return to Main Menu\ \DM\ \BMC\1\VELOCITY LENGTH\ \BMC\2\VELOCITY SPEED\ \BMC\3\VELOCITY VELOCITY\ \BMC\4\VELOCITY VOCAB\ \BMC\5\VELOCITY REVIMore on This Topic\ \SMO\Leave Force & Motion\ \DM\ \BMC\1\VELOCITY\ \BMC\2\ACCELERATION\ \BMC\3\FRICTION\ \BMC\4\EXPERIMENT\ \BMC\5\VOCAB\ \BMC\6\REVIEW\ \BMC\7\MORE\ \BS\NETWORK\ \L\VELOCITY\ \SM\ \SMT\Velocity & Vectors\ \SMO\Length andINCLTESTBFINCLIN1MV5OLDScientists have arbitrarily defined the meter as the unit of length and distance. If you stretch your arm out to the side and face forward, the distance from the tip of your nose to the tip of your fingers is probably about one meter.\ \DTN\3@``0 UUU 000```@ @```000 TPP@@6CJ   * >1111%<`*** xcxUUU pgp*** pkUUUpX | ((/m***ppUUUl@***@i@ UUUxgx ***8`a`8 UUU8`[`8***p@Q@pUUU 0`@E@`0*( *** (* 0``@UUT~WqgA`; ((* )  ;UUUEEEeee555 000```@@#@@@ @@`00 # 00`@@ (*( @`0 ; 0`@ UUU@p K p@ ***`8 U 8`UUU`<]g`}==s ~jjjz:::>.../+++ * #xx0gx`pp88?uU#yx zj*&}uUB?~z*D}UF~j *I?u UKzj*M}uUP?~z*R}@``ppxx\\ZJ*'WWUWxC~j*j*~j*%RJ*+?|auUUU_ U?uURJ)%U_~p?zj*?;;?:?*0`@zj*-u ./?~Gw"W FFff6~~>}uUW} UW_|a~z*j00 fccaa``?~z*+>jzz::}Uu]WULU ?~zJ* }HU~jE*?uCUzj@*}u>U?~z<*}:U* * ~j7*QT?u5U  zj2*UQA}u0U *  ?~z.*P@@@``pp8xq?},U*0xx|a}Ux@Llf63p>p?~j *|` @@xCu U?~p 3MMl`x`|azj* x@ >p?}uU |`@@``pp88?~z*}UU)~,??*%qS`}u@~><` x@9yY~FF~p |`  pq`AOx`< ?~p @@ x@ |`3?~p6fF x@"|< //{k+}uU @@``pp88xC~z\ \DTN\ Distance and time are basic physical quantities. They are basic characteristics of the universe, not the creations of people. The measurement of these quantities, however, requires some arbitrary units of measurement.\ \DTN\ ves, you can ask how far did it move?\ \DTN\ In physics, you measure this distance in meters.\ \DCW\ \DTN\ You can also ask how long did it take for the object to move?\ \DTN\ In physics, you measure this time in seconds.\ \DCWucations, Inc.\ \XDEL\90\ \L\FORCE\ \SM\ \SMT\1: FORCE & MOTION\ \SMO\Velocity and Vectors\ \SMO\Acceleration and Force\ \SMO\Friction and Vectors\ \SMO\Experiment with Falling Bodies\ \SMO\Vocabulary Review\ \SMO\Review Questions\ \SMO\Mon Feb 4 13:03:34 1991 \DCW\ \SCH\2\\ \SCT\0\4\ \DTCNW\FORCE\ \DTCNW\ & MOTION \ \SCH\1\\ \SCT\6\11\ \DTCNW\Robert Stickgold, Ph.D.\ \SCT\8\16\ \DTCNW\Copyright (c) 1990\ \SCT\8\17\ \DTCNW\Queue, Inc.\ \SCT\4\18\ \DTCNW\Intentional Ed Similarly, they have defined the second as the standard unit of time. If you listen to your heartbeat, the time interval between beats is about one second.\ \DTN\ The unit meters is usually abbreviated m. The unit seconds is usually abbreviated s.\ \DCW\ \DTNNW\ On your monitor, this ball is moving about two-tenths of a meter, or 0.2 m.\ \SFF\0\\ \SA\ \DFA\BALL1\268\8\ \SAS\50\1\1\-1\ \SAL\26\-10\0\0\0\ \DA\ \DT\\ \DTN\ The dot on the letter  i  is about t\0\0\ \DA\ \GCR\ \XERASE\248\5\267\20\ \DCW\ \DTNNW\ Now the ball is moving twice as fast as it was before.\ \XDEL\3\ \SA\ \DFA\BALL1\8\8\ \SAS\50\1\1\-1\ \SAL\13\20\0\0\0\ \DA\ \GCR\ \SFF\0\\ \SA\ \DFA\BALL1\272\8\ \SAS\50\1\1\-1\ \SALs\ \DTNW\ \ \DT\ 2 s\ \SCH\1\1\ \DCW\ \DTNNW\ It takes the ball the same amount of time to move in each direction. This means that the ball's speed is the same in each case.\ \SFF\0\\ \SA\ \DFA\BALL1\8\8\ \SAS\50\1\1\-1\ \SAL\26\10\0\DA\ \DTNNW\ To get the ball's speed, we divided the distance it moved by the time it took:\ \SCH\2\3\ \DTNNW\ 0.2 m\ \DTNW\ ------ = 0.1 m/s\ \DT\ 2 s\ \SCH\1\1\ \DCW\ \DTNNW\Or:\ \SCH\2\3\ \DTNNW\ 200 mm \ \DTNW\------- = 100 mm/how fast the ball is moving. In this case, the speed is measured in meters per second.\ \DCW\ \DTNNW\ To determine a speed, you must divide a distance by a time.\ \SFF\0\\ \SA\ \DFA\BALL1\268\8\ \SAS\50\1\1\-1\ \SAL\26\-10\0\0\0\ fast does an object move?\ \DCW\ \SWT\\\6\\ \DTNNW\ Assume that this ball moves about two-tenths of a meter in about 2 seconds.\ \SA\ \DFA\BALL1\8\8\ \SAS\50\1\1\-1\ \SAL\26\10\0\0\0\ \DA\ \DTN\ The speed of the ball is a measure of  and "kilo" means one thousand.\ \SQANS\1000\ \SQLAST\No, there 1000 meters in a kilometer.\ \SQA\T\=\1000\Right, "kilometer" means one thousand meters.\ \DQ\ \SQSEND\ \BESC\ \L\VELOCITY SPEED\ \SCH\2\2\ \DTCNW\Speed\ \SCH\1\1\ \DTN\ How ond.\ \SQANS\1000\ \SQLAST\No, there are 1000 ms in a second.\ \SQA\T\=\1000\Right. "milli" means one-thousandth.\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQM\2\ \SQF\S\4\ \SQT\1 km is equal to ____ m?\ \SQH\Remember that "milli" means one-thousandtheter.\ \SQANS\1000\ \SQLAST\No, there are 1000 mm in a meter.\ \SQA\T\=\1000\Right. "milli" means one-thousandth.\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQM\2\ \SQF\S\4\ \SQT\1 s is equal to ____ ms?\ \SQH\Remember, there are a thousand ms in a sec\SQANS\200\ \SQLAST\No, there are 200 ms in 0.2 s. (0.2 x 1000 = 200 ms).\ \SQA\T\=\200\Right. (0.2 x 1000 = 200 ms).\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQM\2\ \SQF\S\4\ \SQT\1 m is equal to ____ mm?\ \SQH\Remember, there are a thousand mm in a m\ \SQA\T\=\200\Right. (0.2 x 1000 = 200 mm).\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQF\S\4\ \SQM\3\ \SQT\0.2 s is equal to ____ ms.\ \SQH\Remember, there are 1000 ms in a second.\ \SQH\If there are 1000 ms in a second, there are 100 ms in 0.1 s.\ Q\1\Y\ \SQF\S\4\ \SQM\3\ \SQT\0.2 m is equal to ____ mm.\ \SQH\Remember, there are 1000 mm in a meter.\ \SQH\If there are 1000 mm in a meter, there are 100 mm in 0.1 m.\ \SQANS\200\ \SQLAST\No, there are 200 mm in 0.2 m. (0.2 x 1000 = 200 mm). Assume that this ball moves about two-tenths of a meter in about 2 seconds.\ \SFF\0\\ \SA\ \DFA\BALL1\8\8\ \SAS\50\1\1\-1\ \SAL\26\10\0\0\0\ \DA\ \SFF\0\\ \SA\ \DFA\BALL1\268\8\ \SAS\50\1\1\-1\ \SAL\26\-10\0\0\0\ \DA\ \DCW\ \SQS\1\Y\ \Ss = 0.001 s\ \DTN\ Similarly, when working with large quantities of meters or seconds, scientists add the prefix kilo- to the unit to indicate thousands of units. 1 kilometer = 1 km = 1,000 m 1 kilosecond = 1 ks = 1,000 s\ \DCW\ \DTNNW\wo-thousandths of a meter, or 0.002 m.\ \DTN\ When working with small fractions of a meter or a second, scientists add the prefix milli- to the unit to indicate one-thousandths of the unit. 1 millimeter = 1 mm = 0.001 m 1 millisecond = 1 m\13\-20\0\0\0\ \DA\ \SQS\1\Y\ \DCW\ \SQ\1\Y\ \SQM\3\ \SQF\S\1\ \SQT\If the ball moved 200 mm in 2 s and is now going twice as fast, how long does it take the ball to move the 200 mm? a. 0.5 s c. 2 s e. 20 s b. 1.0 s d. 4 s f. 200 s\ \SQH\The ball was initially going 100 mm/s. Think about how fast it is going now.\ \SQH\If it's moving twice as fast, it should take only half as long.\ \SQANS\b\ \SQLAST\No, since it's moving twice as fast, it will take only half aselocity mean the same thing. But physicists use these terms in different ways.\ \DCW\ \SWT\\\9\\ \DTNNW\ A ball moves across the screen in each direction at the same speed of 200 mm/s.\ \SA\ \DFA\BALL1\8\5\ \SAS\50\1\1\-1\ \SAL\13\20\0\0\0 taken.\ \XPLOT\43\91\ \XDRAW\200\-82\ \DTN\ This block moved a total of 8 m in 4 s, for an average speed of 2 m/s.\ \BESC\ \L\VELOCITY VELOCITY\ \SCH\2\2\ \DTCNW\Velocity & Vectors\ \SCH\1\1\ \DTN\ In common usage, the words speed and v a speed of 1 m/s.\ \SCH\\3\ \XPLOT\113\84\ \XDRAW\81\-33\ \SCH\\1\ \DTN\ At 2 sec, the block is moving at a speed of 2 m/s.\ \DCW\ \DTN\ The average speed of the block is obtained by dividing the total distance traveled by the total timeck at any given moment is the slope of the curve at that point.\ \XPLOT\56\94\ \XDRAW\83\-17\ \DTN\ You can measure this slope by drawing a line tangent to the curve at that point and measuring its slope.\ \DTN\ At 1 s, the block is moving attance traveled by a block of wood sliding down an inclined plane.\ \DTN\ As it slides down, its speed increases. This is seen in the increasing slope of the curve as it moves to the right.\ \DCW\ \DTN\ The instantaneous speed of the bloVT\meters\ \SGYMAX\8\ \SGXMAX\4\ \SGVL\2\0\2\ \SLGHL\0\1\ \SLGLX\0\.25\.50\.75\1\1.25\1.50\1.75\2\2.25\2.50\2.75\3\3.25\3.50\3.75\4\ \SLGLY\0\0\.1\.3\.5\.8\1.1\1.5\2\2.5\3.1\3.8\4.5\5.3\6.1\7.0\8.0\ \DLG\ \SWT\M\M\15\M\ \DTN\ This shows the disow far the ball moved at each moment during its travel at the faster speed. Its speed was twice as great and the slope of the line is therefore twice as great.\ \SWS\1\ \DCW\ \SWT\2\35\M\13\ \SLG\1\ \SLGL\1\2\\0\ \SGHT\seconds\ \SGDF\2\1\ \SG\M\M\M\10\ \SLG\2\ \SLGL\1\2\\0\ \SLGL\2\1\\0\ \SGHT\sec\ \SGDF\1\1\ \SGVT\mm\ \SGYMAX\200\ \SGXMAX\3\ \SGVL\3\0\100\ \SLGHL\0\1\ \SLGLX\0\2\ \SLGLY\0\200\ \SLGLX\0\1\ \SLGLY\0\200\ \DLG\ \SWT\M\M\12\M\ \DCW\ \DTN\ The new line shows h200\ \DLG\ \SWT\\\12\M\ \DTN\ This graph shows how far the ball has moved at each moment in its travel at the slower speed. The ball moved 200 mm in 2 s.\ \DTN\ The slope of the line gives the speed of the ball.\ \SWS\1\ \DCW\ \SWT00 mm)/(200 mm/s) = 5 s.\ \SQA\T\=\c\Right! (1000 mm)/(200 mm) = 5\ \DQ\ \SQSEND\ \SWS\1\ \DCW\ \SWT\M\M\M\10\ \SLG\1\ \SLGL\1\2\\0\ \SGHT\sec\ \SGDF\1\1\ \SGVT\mm\ \SGYMAX\200\ \SGXMAX\3\ \SGVL\3\0\100\ \SLGHL\0\1\ \SLGLX\0\2\ \SLGLY\0\s c. 5 s e. 20 s b. 2 s d. 10 s f. 50 s\ \SQH\Remember that 1 meter = 1000 mm.\ \SQH\Think - if it goes 200 mm/s, then to determine how many seconds it needs to go 1000 mm, you must divide 1000/200.\ \SQANS\c\ \SQLAST\No, (10m/s, twice the original speed of 100 mm/s.\ \SQA\T\=\d\Right! 2 x 100 mm/s = 200 mm/s\ \SQA\F\=\b\No, it's going faster now.\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQF\S\1\ \SQT\At 200 mm/s, how long would it take the ball to move 1 meter? a. 1 and now is moving twice as fast, what is the new speed of the ball? a. 10 mm/s c. 100 mm/s e. 400 mm/s b. 50 mm/s d. 200 mm/s f. 800 mm/s\ \SQH\The old speed was 200 mm per 2 seconds, or 100 mm/s.\ \SQANS\d\ \SQLAST\No, the new speed is 200 m long, or 1 s.\ \SQA\T\=\b\Right, if it's moving twice as fast, it will take only half as long.\ \SQA\F\=\d\No, if it's going faster, it will take LESS time.\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQF\S\1\ \SQM\3\ \SQT\If the ball was moving 100 mm/s\ \SAL\13\-20\0\0\0\ \DA\ \DTN\ However, the velocities of the ball on the two legs of the trip are not the same.\ \DTN\ The ball first moves with a velocity of 200 mm/s to the right and then with a velocity of 200 mm/s to the left.\ \DTN\ A velocity is a speed and a direction.\ \SWS\1\ \DCW\ \SWT\\\16\\ \DTNW\ A ball travels a circular path at a constant speed.\ \SA\ \DFA\BALL1\130\0\ \SAS\30\\1\3\ \SAL\10\11\-1\1\1\ \SAL\10\-2\-1\10\-1\ \SAL\10\-11\1\-1\-1\ \SAL\1\ \SWT\M\M\17\M\ \DFB\VECTORS4\56\0\ \XPLOT\97\38\ \XDRAW\8\-8\0\3\0\-3\-3\0\ \XPLOT\182\30\ \XDRAW\8\8\0\-3\0\3\-3\0\ \XPLOT\194\90\ \XDRAW\-8\8\0\-3\0\3\3\0\ \XPLOT\104\98\ \XDRAW\-8\-8\3\0\-3\0\0\3\ \SWT\\\17\\ \DTN\ A ball moves around a direction in which the object is moving.\ \SQANS\down\ \SQLAST\No, the arrow points down, so the object must be moving down.\ \SQA\T\CONTAINS\down\Right. The arrow points down, so the object is moving down.\ \DQ\ \BQESC\ \SQSEND\ \SWS\1\ \DCW3\ \DF\VECTORS3\1\1\1\0\ \SWT\\\R2\\1\ \DCW\ \SQ\1\Y\ \SQF\S\6\ \SQT\This vector represents the velocity of an object moving in what direction?\ \SQH\Enter UP, DOWN, LEFT, or RIGHT.\ \SQH\Remember that the direction of the arrow indicates therrow indicates the direction in which the object is moving.\ \SQANS\up\ \SQLAST\No, the arrow points up, so the object must be moving up.\ \SQA\T\CONTAINS\up\Right. The arrow points up, so the object is moving up.\ \DQ\ \BQESC\ \SWS\1\ \SFF\0\SWS\1\ \SQS\1\Y\ \SFF\0\\ \DF\VECTORS2\1\1\1\0\ \SWT\\\R2\\1\ \SQ\1\Y\ \SQF\S\5\ \SQT\This vector represents the velocity of an object moving in what direction?\ \SQH\Enter UP, DOWN, LEFT, or RIGHT.\ \SQH\Remember that the direction of the aA\T\CONTAINS\r\Correct. The fact that the arrow points to the right indicates that the ball is moving to the right.\ \SQA\T\CONTAINS\l\No, the fact that the arrow points to the right indicates that the ball is moving to the right.\ \DQ\ \BQESC\ \Does the lower vector represent the velocity of the ball moving to the left or to the right?\ \SQH\Enter left or right.\ \SQANS\right\ \SQLAST\No, the fact that the arrow points to the right indicates that the ball is moving to the right.\ \SQat the arrow points to the left indicates that the ball is moving to the left.\ \SQA\T\CONTAINS\r\No, the fact that the arrow points to the left indicates that the ball is moving to the left.\ \DQ\ \BQESC\ \DCW\ \SQ\1\N\ \SQM\2\ \SQF\S\5\ \SQT\y of the ball moving to the left, or moving to the right?\ \SQH\Enter left or right.\ \SQANS\L\left\ \SQLAST\No, the fact that the arrow points to the left indicates that the ball is moving to the left.\ \SQA\T\CONTAINS\l\Correct. The fact thf the object.\ \DTN\ The direction in which the arrow is pointing gives the direction of the vector and indicates the direction in which the object was moving.\ \DCW\ \SQ\1\N\ \SQM\2\ \SQF\S\5\ \SQT\Does the upper vector represent the velocit A vector consists of a magnitude and a direction.\ \SWS\1\ \DF\VECTORS1\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ A velocity vector is represented by an arrow. The length of the arrow gives the magnitude of the vector and represents the speed oistance it moves each second is constant, its direction is constantly changing.\ \DQ\ \DCW\ \SWT\\\8\\ \DTN\ Whereas a speed can be represented simply by a number, a velocity cannot.\ \DTN\ Physicists represent velocities by vectors. the distance it moves each second is constant, its direction is constantly changing.\ \SQA\T\CONTAINS\y\No! Although the distance it moves each second is constant, its direction is constantly changing.\ \SQA\T\CONTAINS\n\Right! Although the d0\2\1\-10\1\ \SAL\10\11\-1\1\1\ \SAL\10\-2\-1\10\-1\ \SAL\10\-11\1\-1\-1\ \SAL\10\2\1\-10\1\ \DA\ \DCW\ \SWT\\\15\\ \SQ\1\N\ \SQM\2\ \SQF\S\3\\QH\ \SQT\Was the ball's velocity constant?\ \SQH\Enter Y or N\ \SQANS\no\ \SQLAST\No! Although circle at a constant speed. But its velocity is constantly changing.\ \DTN\ While the magnitude of the velocity remains constant, its direction changes constantly.\ \DTN\ The instantaneous velocity of the ball at a given moment is its instantaneous speed and the direction it is moving at that moment.\ \SFF\0\1\ \DCW\ \DFB\V5OLD\56\0\ \SWT\\\17\\ \DTN\ These arrows indicate the instantaneous velocity of the ball at various moments as it traveled around the circle.\ \ \DFA\BALL1\8\1\ \SAS\40\0\1\-1\ \SAL\18\0\0\0\1\ \DA\ \GCR\ \DCW\ \DTN\ You can see the ball speeding up more clearly if we take stroboscopic pictures of it falling.\ \XERASE\10\152\25\164\ \SA\ \DFA\BALL1\8\0\ \SAS\40\0\0\-1\ \SAL\18\0 speed gradually increases.\ \SFF\3\\ \SA\ \DFA\BALL1\8\1\ \SAS\40\0\1\-1\ \SAL\18\0\0\0\1\ \DA\ \DTNNW\ You might imagine that this is a picture of a ball falling from a height of 50 m to the ground below in 3 s.\ \XERASE\10\152\25\164\ \SA The speed or velocity of an object tells you how rapidly its position is changing.\ \DTN\ The acceleration of an object tells you how rapidly its speed or velocity is changing.\ \DCW\ \SWT\8\\\\ \DTNNW\ When this ball falls, its \ \BMC\2\ACCEL CONSTANT\ \BMC\3\ACCEL NEWT1\ \BMC\4\ACCEL NEWT2\ \BMC\5\ACCEL VOCAB\ \BMC\6\ACCEL REVIEW\ \BL\FORCE\ \L\ACCEL ACCEL\ \SCH\2\2\ \DTCNW\Acceleration\ \SCH\1\1\ \DTN\ The position of an object tells you where it is located. ERATION\ \SM\ \SMT\Acceleration and Force\ \SMO\Acceleration\ \SMO\Constant Acceleration\ \SMO\Newton's First Law\ \SMO\Newton's Second Law\ \SMO\Vocabulary Review\ \SMO\Review Questions\ \SMO\Return to Main Menu\ \DM\ \BMC\1\ACCEL ACCELt's moving upward, as indicated by the arrow at point D.\ \SQA\T\CONTAINS\up\Correct! The arrow at Point D shows the direction of this vector.\ \DQ\ \SQSEND\ \BESC\ \L\VELOCITY VOCAB\ \RVR\1\ \BESC\ \L\VELOCITY REVIEW\ \RRQ\1\ \BESC\ \L\ACCEL direction of the velocity vector at point D? \ \SQH\No, remember that the velocity vector is represented by the arrow with its end at point D.\ \SQH\No, think about the direction that the arrow at point D is pointing.\ \SQANS\up\ \SQLAST\No, iNS\left\ \SQLAST\No, it's moving to the left, as indicated by the arrow at point C.\ \SQA\T\CONTAINS\left\Correct! The arrow at Point C shows the direction of this vector.\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQM\3\ \SQF\S\6\\NQAH\ \SQT\What is theAH\ \SQT\What is the direction of the velocity vector at point C? \ \SQH\No, remember that the velocity vector is represented by the arrow with its end at point C.\ \SQH\No, think about the direction that the arrow at point C is pointing.\ \SQAt point B is pointing.\ \SQANS\down\ \SQLAST\No, it's moving down, as indicated by the arrow at point B.\ \SQA\T\CONTAINS\down\Correct! The arrow at Point B shows the direction of this vector.\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQM\3\ \SQF\S\6\\NQ \SQT\What is the direction of the ball's velocity vector at point B? \ \SQM\3\ \SQF\S\6\\NQAH\ \SQH\No, remember that the velocity vector is represented by the arrow with its end at point B.\ \SQH\No, think about the direction that the arrow aion that the arrow at point A is pointing.\ \SQANS\right\ \SQLAST\No, it's moving to the right, as indicated by the arrow at point A.\ \SQA\T\CONTAINS\right\Correct! The arrow at Point A shows the direction of this vector.\ \DQ\ \DCW\ \SQ\1\Y\ DCW\ \SQ\1\Y\ \SQM\3\ \SQF\S\6\\NQAH\ \SQT\What is the direction of the velocity vector for the ball at point A? \ \SQH\No, remember that the velocity vector is represented by the arrow with its end at point A.\ \SQH\No, think about the direct\DTN\ In each case, the tail end of the vector arrow is placed at the position of the ball where the velocity is being measured.\ \SQS\1\Y\ \SWS\1\ \DCW\ \SWT\M\M\17\M\ \SWS\1\ \DCW\ \SWT\M\M\17\M\ \SFF\0\\ \DFB\VECTORS5\56\0\ \DFL\1\4\ \\0\0\1\ \DA\ \DTN\ The distance between frozen pictures is an indication of the speed of the ball at each point.\ \DTN\ If it moves 50 m in 3 s, its average speed is 16.7 m/s. This can be seen from a graph of position versus time.\ \SWS\1\ \SFF\0\\ \DF\ACCEL1\1\1\1\0\ \SWT\M\M\17\M\ \DT\ As time passes, the curve becomes steeper and steeper. This increase in slope reflects an increase in velocity.\ \XPLOT\60\4\ \XDRAW\176\100\ \DTN\ The line connecting the initial d\Right, it increased from 10 to 20 m/s.\ \SQA\F\=\e\No, that's the total velocity at two seconds. You need the increase in velocity from t = 1 to t = 2.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\2\ \SQF\S\1\\QH\ \SQT\What is the velocity of the ball ang the second second? a. 0 m/s c. 5 m/s e. 20 m/s b. 1 m/s d. 10 m/s f. 30 m/s\ \SQH\The velocity was 10 m/s at t = 1 s and 20 m/s at t = 2 s.\ \SQANS\d\ \SQLAST\No, the speed increased by 10 m/s, from 10 m/s to 20 m/s.\ \SQA\T\=\ity graph, for t = 2 s\ \SQANS\e\ \SQLAST\No, the velocity was 20 m/s.\ \SQA\T\=\e\Right. The graph shows the velocity was 20 m/s at 2 s.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\2\ \SQF\S\1\\QH\ \SQT\How much does the velocity of the ball change duri\ \SQA\T\=\d\Right, it increased from 0 to 10 m/s.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\2\ \SQF\S\1\\QH\ \SQT\What is the velocity of the ball at 2 s? a. 0 m/s c. 5 m/s e. 20 m/s b. 1 m/s d. 10 m/s f. 30 m/s\ \SQH\Look at the velocl change during the first second? a. 0 m/s c. 5 m/s e. 20 m/s b. 1 m/s d. 10 m/s f. 30 m/s\ \SQH\The velocity was 0 m/s at t = 0 s, and 10 m/s at t = 1 s.\ \SQANS\d\ \SQLAST\No, the speed increased by 10 m/s, from 0 m/s to 10 m/s. at the velocity graph for t = 1 s\ \SQANS\d\ \SQLAST\No, the velocity was 10 m/s.\ \SQA\T\=\d\Right. The graph shows that the speed was 10 m/s at 1 s.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\2\ \SQF\S\1\\QH\ \SQT\How much does the velocity of the balng at all.\ \SQA\T\=\a\Right. It wasn't moving at all.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\2\ \SQF\S\1\\QH\ \SQT\What is the velocity of the falling ball at 1 s? a. 0 m/s c. 5 m/s e. 20 m/s b. 1 m/s d. 10 m/s f. 30 m/s\ \SQH\LookQM\2\ \SQF\S\1\\QH\ \SQT\What is the velocity of the ball at time zero? a. 0 m/s c. 5 m/s e. 20 m/s b. 1 m/s d. 10 m/s f. 30 m/s\ \SQH\Look at the velocity graph for t = 0 s\ \SQANS\a\ \SQLAST\No, the velocity was 0 m/s; it wasn't movi constant rate.\ \DCW\ \DTN\ Since the height of the curve increases by 10 m/s every second, the slope of the line is 10 m/s per second, or 10 meters per second per second.\ \DCW\ \SQS\2\Y\ \SFF\0\\ \DF\ACCEL3\1\0\1\0\ \SWT\\\R2\\1\ \SQ\ \Sn of time on this same graph.\ \SFF\0\\ \DF\ACCEL3\1\1\1\0\ \SWT\\\R2\\ \DTN\ The curve on the left shows the velocity of the ball as a function of time.\ \DTN\ The curve forms a straight line, indicating that the velocity is increasing at a0\1\-16\0\ \GCR\ \SWT\M\M\17\M\ \DCW\ \SWT\M\M\18\M\ \SCH\1\1\ \DTN\ This acceleration is in units of meters per second per second, or meters per second squared.\ \DTN\ We can show this by plotting the instantaneous velocity as a functioSWT\M\M\17\M\ \SCH\2\3\ \DCW\ \SWT\M\M\17\M\ \DTNNW\ 30 m/s 10 m \ \DTNW\ 3 s s2\ \XPLOT\27\160\ \XDRAW\84\\0\1\-84\0\ \XPLOT\165\160\ \XDRAW\63\0\0\1\-63\0\ \XPLOT\129\156\ \XDRAW\16\0\0\-1\-16\0\ \XPLOT\129\164\ \XDRAW\16\0\0 and 30 m/s, respectively.\ \DTN\ At time zero, the instantaneous velocity is actually zero.\ \DCW\ \DTN\ To calculate the average acceleration, take the change in velocity over the entire period, 30 m/s, and divide it by the time, 3 s.\ \and final position of the curve indicates an average velocity of 16.7 m/s.\ \SCH\1\3\ \XPLOT\93\4\ \XDRAW\57\19\ \XPLOT\212\62\ \XDRAW\28\33\ \SCH\1\1\ \DTN\ The two tangent lines show that the instantaneous velocities at 1 s and 3 s are 1t 3 s? a. 0 m/s c. 5 m/s e. 20 m/s b. 1 m/s d. 10 m/s f. 30 m/s\ \SQANS\f\ \SQLAST\No, the velocity was 30 m/s.\ \SQH\Look at the graph at t = 3 s.\ \SQA\T\=\f\Right. The graph shows that the velocity was 30 m/s at 3 s.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\2\ \SQF\S\1\\QH\ \SQT\How much does the velocity of the ball change during the third second? a. 0 m/s c. 5 m/s e. 20 m/s b. 1 m/s d. 10 m/s f. 30 m/s\ \SQH\The velocity was 20 m/s at t = 2 s and 30 m/s ue as a function of time as well.\ \DCW\ \XERASE\0\0\279\47\ \DFB\CRAZYV\14\0\ \DFB\CRAZYP\0\48\ \SWT\\\14\\ \DCW\ \DTN\ Notice that the velocity is positive when the ball is moving toward the right and negative when it's moving toward the of the ball during its movement.\ \DTN\ The curve goes up and down as the ball moves to the right and then back to the left.\ \DTN\ At every point the slope of the curve gives the instantaneous velocity of the ball. We can plot this valA\BALL1\4\17\ \SAS\50\\1\3\ \SAL\10\0\1\0\0\ \SAL\10\10\0\0\0\ \SAL\10\10\-3\0\0\ \SAL\5\-20\8\0\0\ \SAL\12\20\0\0\0\ \DA\ \SWT\\\10\\ \DTN\ Got that?\ \SWS\1\ \DCW\ \SFF\0\\ \DFB\CRAZYP\0\0\ \SWT\M\M\8\M\ \DTN\ This shows the positionEXT\30\180\33\ \XTEXT\40\236\33\ \SA\ \DFA\BALL1\4\17\ \SAS\50\\1\3\ \SAL\10\0\1\0\0\ \SAL\10\10\0\0\0\ \SAL\10\10\-3\0\0\ \SAL\5\-20\8\0\0\ \SAL\12\20\0\0\0\ \DA\ \SWT\\\10\\ \DTNNW\ Let's see that again.\ \XERASE\237\17\250\27\ \SA\ \DF very smoothly.\ \DTN\ This need not be the case.\ \SWS\1\ \DCW\ \SWT\M\M\17\M\ \SWT\M\M\17\M\ \XTEXT\........................................\0\24\ \XPLOT\0\28\ \XDRAW\279\0\0\1\-279\0\ \XTEXT\0\13\33\ \XTEXT\10\68\33\ \XTEXT\20\124\33\ \XTT\\\R2\\1\ \DCW\ \DTN\ Interestingly, the acceleration is greatest at the two ends, where the velocity is smallest. The acceleration is smallest where the velocity is greatest.\ \DTN\ In this example, the acceleration is still changingo at each end.\ \DTN\ Notice that the slope of the velocity curve is far from constant.\ \DTN\ This means that the velocity is not changing at a constant rate. The acceleration of the ball is constantly changing.\ \DF\HARMONY3\1\1\1\0\ \SWCW\ \SFF\0\\ \DF\HARMONY1\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ This graph shows the position as a function of time.\ \DF\HARMONY2\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ The velocity of the ball is greatest in the middle of its path. Of course, it must be zerare constantly changing.\ \SA\ \DFA\BALL1\80\12\ \SAS\90\\1\\ \SAL\4\1\3\0\0\ \SAL\4\18\0\0\0\ \SAL\4\10\-3\0\0\ \SAL\4\-1\-3\0\0\ \SAL\4\-18\0\0\0\ \SAL\4\-10\3\0\0\ \SAL\4\1\3\0\0\ \SAL\4\18\0\0\0\ \SAL\4\10\-3\0\0\ \DA\ \GCR\ \SWS\1\ \DPLOT\148\25\ \XDRAW\0\2\1\0\0\-2\ \XPLOT\86\25\ \XDRAW\0\2\1\0\0\-2\ \XPLOT\208\25\ \XDRAW\0\2\1\0\0\-2\ \XTEXT\0\84\29\ \XTEXT\1\146\29\ \XTEXT\2\206\29\ \SWT\\\7\\ \DTNNW\ In this example, both the velocity and the acceleration of the ball \No, this is its velocity at 3 s.\ \SQA\F\=\f\No, at this rate it would reach a velocity of 30 m/s in just 1 s.\ \DQ\ \SQSEND\ \DCW\ \DTN\ Acceleration need not always be constant.\ \SWS\1\ \DCW\ \XPLOT\83\23\ \XDRAW\130\0\0\1\-130\0\ \Xond, for a rate of 10 m/s/s (or 10 m/s2).\ \SQA\T\=\e\Right. It accelerated 10 m/s each second, a rate of 10 m/s/s (10 m/s2).\ \SQA\F\IS CONTAINED IN\ab\No, this is a distance!\ \SQA\F\=\c\No, this is its velocity at 1 s.\ \SQA\F\=\dhat is the acceleration of the ball during these three seconds? a. 10 m c. 10 m/s e. 10 m/s2 b. 50 m d. 30 m/s f. 30 m/s2\ \SQH\How much did the ball speed up each second?\ \SQANS\e\ \SQLAST\No, it accelerated by 10 m/s every secat t = 3 s.\ \SQANS\d\ \SQLAST\No, the speed increased by 10 m/s, from 20 m/s to 30 m/s.\ \SQA\T\=\d\Right, it increased from 20 to 30 m/s.\ \SQA\F\=\f\No, that's the velocity at t = 3 s.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\4\ \SQF\S\1\\QH\ \SQT\Wleft.\ \DTN\ When the velocity is changing, the ball is accelerating.\ \SWS\1\ \DCW\ \DFB\CRAZYA\0\0\ \DFB\CRAZYV\14\42\ \DFB\CRAZYP\0\90\ \SWT\M\M\19\M\ \DCW\ \DTN\ The faster the |C velocity changes|C, the more the ball ~ is |C accelerating|C.\ \DTN\ The slower the velocity changes, the less the ball is accelerating.\ \DTN\ When the velocity is constant, acceleration is zero.\ \DTN\ Notice that when the ball is speeding up, the acceleration is greater than zero. When the acceleration is positive, the velocity curve moves upward.\ \DTN\ When the acceleration is negative, the ball moves less rapidly toward the right, or more rapidly toward the left.\ \DTN\ When the acceleration is negative, the veloc\DTN\ When the velocity is negative, the ball moves toward the left and the position curve moves downward.\ \DCW\ \DTN\ When the acceleration is positive, the ball moves more rapidly toward the right or less rapidly to the left.\ \DTN\ me zero.\ \DQ\ \SQSEND\ \SWS\1\ \DCW\ \DTN\ Notice that both the velocity and acceleration can be either positive or negative.\ \DTN\ When the velocity is positive, the ball moves toward the right and the position curve moves upward.\ = 0 s) is ____ m/s2. \ \SQH\Look at the acceleration curve for t = 0 s.\ \SQANS\50\ \SQLAST\No, the acceleration curve shows a value of 50 m/s2 at time zero.\ \SQA\T\=\50\Right. The acceleration curve shows a value of 50 m/s2 at tilocity graph for t = 0 s.\ \SQANS\0\ \SQLAST\No, it isn't moving at all at time zero.\ \SQA\T\=\0\Right. It isn't moving at all at t = 0.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\3\\NQAH\ \SQT\The acceleration of the ball at this time (ts lowest point at t = 0 s.\ \SQA\T\=\0 s\Right. The position graph is at its lowest point at t = 0 s.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\3\\NQAH\ \SQT\The speed of the ball at this time (t = 0 s) is ____ m/s. \ \SQH\Look at the veT\The ball is furthest to the left at t = ____ s. \ \SQH\At what time is the position graph at its lowest level?\ \SQANS\0\ \SQLAST\No, the position graph is at its lowest level at time zero.\ \SQA\T\=\0\Right. The position graph is at it fast is it changing?!?\ \SQANS\0\ \SQLAST\No, its acceleration is 0 m/s2. It isn't accelerating at all if its velocity stays constant!\ \SQA\T\=\0\Right. It isn't accelerating at all.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\3\\NQAH\ \SQ=\0\Right. The acceleration curve at t = 2.5 s shows a value of 0 m/s2.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\3\\NQAH\ \SQT\The acceleration at times when the velocity is constant is ____ m/s2. \ \SQH\If the velocity is constant, how3\\NQAH\ \SQT\The acceleration of the ball at this time (t = 2.5 s) is ____ m/s2. \ \SQH\Look at the acceleration curve at t = 2.5 s.\ \SQANS\0\ \SQLAST\No, the curve shows that the rate of acceleration at t = 2.5 s is 0 m/s/s.\ \SQA\T\\ \SQH\Look at the graph of velocity versus time, at t = 2.5 s.\ \SQANS\50\ \SQLAST\No, at t = 2.5 s the velocity is 50 m/s.\ \SQA\T\CONTAINS\50\Right. The velocity graph is at 50 m/s at 2.5 s.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\SQA\T\=\2.5\Right. At 2.5 s the graph is at its highest point.\ \SQA\F\IS CONTAINED IN\2.\No, it's a little later than that!\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\3\\NQAH\ \SQT\The speed of the ball at this time (t = 2.5 s) is ____ m/s. the right at t = ____ s. \ \SQH\Look at the graph of position versus time.\ \SQH\At what time is the position highest?\ \SQANS\2.5\ \SQLAST\No, the ball is furthest to the right at 2.5 s, when the graph of position versus time is highest.\ \\ \DTN\ When the ball is slowing down, the acceleration is less than zero.\ \SWS\1\ \DCW\ \SQS\2\Y\ \SFF\0\\ \DFB\CRAZYA\0\0\ \DFB\CRAZYV\14\42\ \DFB\CRAZYP\0\92\ \SWT\M\M\18\M\ \SQ\ \SQM\3\ \SQF\S\3\\NQAH\ \SQT\The ball is furthest to ity curve moves downward.\ \BESC\ \L\ACCEL CONSTANT\ \SCH\2\2\ \DTCNW\Constant Acceleration\ \SCH\1\1\ \DTN\ When an object falls without hindrance, it accelerates at a constant rate of 9.8 m/s2, or approximately 10 m/s2.\ \DCW\ \SWT\8\M\M\M\ \DTN\ When this ball falls, its speed gradually increases.\ \SA\ \DFA\BALL1\8\0\ \SAS\40\0\1\-1\ \SAL\18\0\0\0\1\ \DA\ \DTN\ You might imagine that this is a picture of a ball falling from a height of 50 m to the ground below in 3 s.\QF\S\5\ \SQT\Its velocity at t = 1 s will be ____ m/s.\ \SQH\Remember that the final speed is the initial speed plus the amount it sped up.\ \SQH\Remember that its initial speed was 10 m/s and it sped up 9.8 m/s each second for 1 second.\ \SQANSnitial velocity.\ \DTN\ This equation is correct in all instances of constant, or uniform, acceleration.\ \SQS\2\Y\ \DCW\ \DTNNW\A ball with an initial velocity of 10 m/s, accelerating at a rate of 9.8 m/s2.\ \SWT\\\R2\\ \SQ\ \SQM\3\ \S(v0).\ \DCW\ \DTNNW\ This gives us the equation:\ \SCH\2\3\ \DTC\v(t) = at + v\ \SCH\1\1\ \SWT\M\M\6\M\ \DTN\ In other words, the velocity at time, t, is equal to the rate of acceleration, a, multiplied by the time, t, plus the id by the time, t.\ \DTN\ If the ball is moving at time t0, this equation will not be correct.\ \DTN\ While the change in velocity can still be calculated as acceleration multiplied by time, this change must be added to the initial velocity r a body moving with constant acceleration, with a velocity (v) equal to 0 m/s when the time (t) is 0, we can say:\ \SCH\2\3\ \DTC\v(t) = at\ \SCH\1\1\ \DTN\ In other words, the velocity at time t is equal to the rate of acceleration, a, multiplie29.4 m/s\ \SQH\Remember, it has been speeding up by 9.8 m/s each second for 1.5 seconds.\ \SQANS\d\ \SQLAST\No, (9.8 m/s2) x (1.5 s) = 14.7 m/s\ \SQA\T\=\d\Right. (9.8 m/s2) x (1.5 s) = 14.7 m/s\ \DQ\ \SQSEND\ \SWS\1\ \DCW\ \DTNNW\ Fom/s\ \SQA\F\=\c\No, this is its speed at 1 s.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\What is the velocity of the falling ball at 1.5 s? a. 0 m/s d. 14.7 m/s b. 2 m/s e. 19.6 m/s c. 9.8 m/s f. s e. 19.6 m/s c. 9.8 m/s f. 29.4 m/s\ \SQH\Remember that it has been accelerating at the rate of 9.8 m/s for two seconds.\ \SQANS\e\ \SQLAST\No, (9.8 m/s2) x (2 s) = 19.6 m/s\ \SQA\T\=\e\Right. (9.8 m/s2) x (2 s) = 19.6 during the one second.\ \SQA\F\IS CONTAINED IN\abc\No, this a rate of acceleration, not a speed.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\What is the velocity of the falling ball at 2 s? a. 0 m/s d. 14.7 m/s b. 2 m/ 1 m/s e. 1 m/s2 c. 9.8 m/s f. 9.8 m/s2\ \SQH\Remember that its speed increases by 9.8 m/s every second.\ \SQANS\c\ \SQLAST\No, the speed at one second will be just 9.8 m/s.\ \SQA\T\=\c\Right, it speeds up by 9.8 m/s t.\ \SWS\1\ \DCW\ \SQS\2\Y\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\Given that the ball is not moving at time zero (t = 0) and that it falls with an acceleration of 9.8 m/s2, what is its velocity at 1 s? a. 0 m/s d. 0 m/s2 b.e ball at each moment during its decent.\ \DCW\ \DTN\ As was mentioned earlier, the acceleration is constant, at:\ \SCH\2\3\ \DTC\9.8 m/s\ \SCH\1\1\ \DTN\ Knowing this, you can calculate the velocity of the ball at any time after the star\-1\ \SAL\18\0\0\0\1\ \DA\ \DTN\ The distance between frozen pictures is an indication of the speed of the ball at that point.\ \SFF\0\\ \DF\ACCEL2\1\1\1\0\ \SWT\M\\R2\\ \DTN\ This graph shows the position, velocity, and acceleration of th \XERASE\10\152\25\165\ \SA\ \DFA\BALL1\8\0\ \SAS\40\0\1\-1\ \SAL\18\0\0\0\1\ \DA\ \DTN\ You can see the ball speeding up more clearly if we take stroboscopic pictures of it falling.\ \XERASE\10\152\25\165\ \SA\ \DFA\BALL1\8\0\ \SAS\40\0\0\19.8\ \SQLAST\No, (10 m/s) + (9.8 m/s2) x (1 s) = 19.8 m/s\ \SQA\T\=\19.8\Right. (10 m/s) + (9.8 m/s2) x (1 s) = 19.8 m/s\ \SQA\T\=\19.80\Right. (10 m/s) + (9.8 m/s2) x (1 s) = 19.8 m/s\ \SQA\F\CONTAINS\20\Please be more accurate.\ \SQA\F\CONTAINS\9.8\Don't forget to include the initial speed.\ \DQ\ \DCW\ \SQ\ \SQM\3\ \SQF\S\5\ \SQT\Its velocity at t = 2.5 s will be ____ m/s.\ \SQH\Remember that its initial speed was 10 m/s and it sped up 9.8 m/s each second for 2.5 seconds. on it?\ \SQANS\yes\ \SQH\Did its velocity change as it rolled to a stop?\ \SQANS\Yes\ \SQLAST\Yes, a force did act. Otherwise, it wouldn't have changed its velocity!\ \SQA\T\CONTAINS\Y\Right! Otherwise it wouldn't have changed its velocity.\ \ \DCW\ \SQS\2\Y\ \SWT\M\M\6\M\ \DTNNW\ Watch this animation.\ \DCW\ \SFF\3\\ \SA\ \DFA\BALL1\0\8\ \SAS\10\\1\3\ \SAL\22\22\-1\0\0\ \DA\ \SWT\\\6\\ \DCW\ \SQ\ \SQM\3\ \SQF\S\7\ \SQT\As this ball rolled to a stop, was any force actingn.\ \SWS\1\ \DCW\ \DTNNW\ Newton's First Law of Motion states this fact:\ \DTN\ If no force is applied to an object, its velocity cannot change. As a result, an object at rest will stay at rest and an object in motion will stay in motion.0\16\253\26\ \SFF\3\\ \SA\ \DFA\BALL1\4\16\ \SAS\10\\1\3\ \SAL\13\0\1\0\0\ \SAL\10\13\0\0\0\ \SAL\13\13\-3\0\0\ \SAL\5\-26\8\0\0\ \SAL\14\14\0\0\0\ \DA\ \DTN\ If there were no forces acting on it, the ball would neither speed up nor slow dow\1\0\0\ \SAL\10\13\0\0\0\ \SAL\13\13\-3\0\0\ \SAL\5\-26\8\0\0\ \SAL\14\14\0\0\0\ \DA\ \DCW\ \DTN\ If this were a real ball rolling along a surface, there would have to be some force acting on it each time it sped up or slowed down.\ \XERASE\24 ball as it moves.\ \XTEXT\........................................\0\24\ \XPLOT\0\28\ \XDRAW\279\0\0\1\-279\0\ \XTEXT\0\13\33\ \XTEXT\10\68\33\ \XTEXT\20\124\33\ \XTEXT\30\180\33\ \XTEXT\40\236\33\ \SA\ \DFA\BALL1\4\16\ \SAS\10\\1\3\ \SAL\13\0SQSEND\ \BESC\ \L\ACCEL NEWT1\ \SCH\2\2\ \DTCNW\Newton's First Law\ \SCH\1\1\ \DTN\ A force is anything that can change the velocity of an object.\ \DTN\ A push or pull is a force.\ \SWS\1\ \DCW\ \SWT\M\M\9\M\ \DTNNW\ Watch this_ m.\ \SQANS\44.1\ \SQLAST\No, (1/2) x (9.8 m/s2) x (3 s)2 = 44.1 m\ \SQA\T\=\44.1\Right. (1/2) x (9.8 m/s2) x (3 s)2 = 44.1 m\ \SQA\T\=\44.10\Right. (1/2) x (9.8 m/s2) x (3 s)2 = 44.1 m\ \SQA\F\=\44\More accuracy, please.\ \DQ\ \=\19.6\Right. (1/2) x (9.8 m/s2) x (2 s)2 = 19.6 m\ \SQA\T\=\19.60\Right. (1/2) x (9.8 m/s2) x (2 s)2 = 19.6 m\ \SQA\F\=\20\More accuracy, please.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\5\ \SQT\When t = 3 s, it will have fallen ___(1/2) x (9.8 m/s2) x (1 s)2 = 4.9 m\ \SQA\F\=\5\More accuracy, please.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\5\ \SQT\When t = 2 s, it will have fallen ____ m.\ \SQANS\19.6\ \SQLAST\No, (1/2) x (9.8 m/s2) x (2 s)2 = 19.6 m\ \SQA\T\ \SQM\3\ \SQF\S\5\ \SQT\At t = 1 s, it will have fallen ____ m.\ \SQH\d = (1/2)at2\ \SQANS\4.9\ \SQLAST\No, (1/2) x (9.8 m/s2) x (1 s)2 = 4.9 m\ \SQA\T\=\4.9\Right. (1/2) x (9.8 m/s2) x (1 s)2 = 4.9 m\ \SQA\T\CONTAINS\4.90\Right. \1\ \DTN\where d is the distance traveled, a is the rate of acceleration and t is the time since the ball was released.\ \DCW\ \SQS\2\Y\ \DTNNW\A ball is dropped at time zero, and then accelerates at a rate of 9.8 m/s2.\ \SWT\\\R2\\ \SQ\ velocity.\ \DQ\ \SQSEND\ \SWS\1\ \DCW\ \DTNNW\In the case of a ball being dropped, the position of the ball at some time, t, can be calculated from the equation:\ \SCH\2\3\ \DTNNW\ 1 \ \DTNW\ d = -at \ \DTNW\ 2\ \SCH\1\ \SQANS\34.5\ \SQLAST\No, (10 m/s) + (9.8 m/s2) x (2.5 s) = 34.5 m/s\ \SQA\T\CONTAINS\34.5\Right. (10 m/s) + (9.8 m/s2) x (2.5 s) = 34.5 m/s\ \SQA\F\CONTAINS\34\More accuracy, please.\ \SQA\F\CONTAINS\24\Don't forget to include the initial \SQA\T\CONTAINS\N\Yes, a force did act! Otherwise it wouldn't have changed its velocity.\ \DQ\ \SWS\1\ \DCW\ \SWT\8\\\\ \DTNNW\Watch this animation.\ \SA\ \DFA\BALL1\8\0\ \SAS\40\0\1\-1\ \SAL\18\0\0\0\1\ \DA\ \SWT\M\M\M\17\ \DCW\ \SWS\1\ \SQ\ \SQM\3\ \SQF\S\7\ \SQT\As this ball fell, was any force acting on it?\ \SQANS\yes\ \SQH\Did its velocity change as it rolled to a stop?\ \SQANS\Yes\ \SQLAST\Yes, a force did act. Otherwise, it wouldn't have changed its velocity!\ \SQ.\ \DTN\ A gram is one-thousandth of a kilogram. The mass of a quarter teaspoon of water is about 1 g.\ \DCW\ \DTNNW\ Newton's Second Law states that the acceleration caused by the application of a force is proportional to the magnitude o the earth.\ \DCW\ \DTN\ Like length and time, mass is a fundamental physical quantity. Scientists measure the mass of an object in kilograms. The unit kilogram is normally abbreviated kg. The mass of 1 liter of water is 1 kgan object is actually a measure of the magnitude of the gravitational force acting on the object.\ \DTN\ Whereas the mass of an object is a property of that object, the weight of an object is a measure of the interaction between that object anderty of all matter. It is a measure of how difficult it is to accelerate the object when a force is applied to it.\ \DTN\ At the surface of the earth, the mass of an object is proportional to the weight of the object. But the weight of twice as massive (and the force is not doubled) it will accelerate at half the rate.\ \XERASE\74\101\109\125\ \SA\ \DFA\BIGBALL5\70\107\ \SAS\300\\1\3\ \SAL\15\0\1\0\-1\ \DA\ \XERASE\176\0\212\20\ \GCR\ \SWS\1\ \DCW\ \DTN\ Mass is a prop as hard, the object will accelerate at twice the rate.\ \SWT\M\M\19\M\ \SFF\3\\ \SA\ \DFA\BIGBALL\70\107\ \SAS\100\\1\3\ \SAL\10\4\2\-4\-2\ \DA\ \XERASE\160\18\195\37\ \GCR\ \DFB\BIGBALL4\70\100\ \SWT\\\19\\ \DCW\ \DT\ If the object isL\15\0\1\0\-1\ \DA\ \XERASE\176\0\212\20\ \GCR\ \DCW\ \DTN\ The rate at which it speeds up will depend on how hard you push and how massive the object is.\ \DCW\ \SFF\0\\ \SWT\\\19\\ \DFB\BIGBALL3\70\93\ \DCW\ \DTN\ If you push twiceWT\\\17\\ \DTN\ If you push on something, it will start to move. If there is no friction to slow it down, the object will continue to speed up as long as you keep pushing.\ \XERASE\74\101\109\125\ \SA\ \DFA\BIGBALL\70\107\ \SAS\200\\1\3\ \SA object causes the object to accelerate at a constant rate, proportional to the magnitude of the force and inversely proportional to the mass of the object.\ \DTN\ This law is rather easily understood intuitively.\ \DCW\ \DFB\BIGBALL2\70\100\ \SDTN\ The direction of the force is the direction in which you try to push the cylinder. It is indicated by the direction in which the arrow is pointing.\ \SWS\1\ \DCW\ \DTNNW\ Newton's Second Law states that:\ \DTN\A force acting on an To move this cylinder, you must apply a force to it.\ \XERASE\72\107\106\125\ \DFB\BIGBALL2\70\100\ \DTN\ The magnitude of the force is a measure of how hard you push on the cylinder. It is indicated here by the length of the arrow.\ \ Force, like velocity, is a vector quantity. A force has both a magnitude and a direction.\ \DCW\ \DFB\BIGBALL1\70\107\ \SWT\\\17\\ \DTN\ Imagine that you are looking down on a large steel cylinder sitting on a flat floor.\ \DTN\ Law\ \SCH\1\1\ \DTN\ Newton's First Law states that the velocity of an object can be altered only as a result of a force acting on the object.\ \DTN\ Newton's Second Law states how a given force will affect the velocity of an object.\ \DTN\A\T\CONTAINS\Y\Right! Otherwise it wouldn't have changed its velocity.\ \SQA\T\CONTAINS\N\Yes, a force did act! Otherwise it wouldn't have changed its velocity.\ \DQ\ \SWS\1\ \SQSEND\ \BESC\ \L\ACCEL NEWT2\ \SCH\2\2\ \DTCNW\Newton's Secondf that force and is inversely proportional to the mass of the object. This can be expressed mathematically as:\ \SCH\2\2\ \DTCNW\a = F/m\ \SCH\1\1\ \DTN\The acceleration equals the force divided by the mass.\ \DCW\ \DTN\ Force is measured in units called newtons. A force of one newton is required to accelerate a one kilogram mass at the rate of one meter per second per second. The unit newton is abbreviated N:\ \SCH\2\2\ \DTN\ 1 m/s = 1 N / 1 kg\ \SQS\2\Y\ \T\An object that accelerates at 2 m/s2 when exposed to a force of 4 N has a mass of ____ kg.\ \SQH\Remember, m = F/a\ \SQANS\2\ \SQLAST\No, (4 N)/(2 m/s2) = 2 kg\ \SQA\T\=\2\Right. (4 N)/(2 m/s2) = 2 kg\ \SQA\F\=\8\No, m = F/a, not F x aSQA\F\=\1000\No, 50 g = 0.05 kg and 20 mm/s2 = 0.02 m/s2\ \SQA\F\=\1\No, 50 g = 0.05 kg and 20 mm/s2 = 0.02 m/s2\ \SQA\F\=\1.0\No, 50 g = 0.05 kg and 20 mm/s2 = 0.02 m/s2\ \DQ\ \SQSEND\ \SQS\2\Y\ \DCW\ \SQ\ \SQM\3\ \SQF\S\3\ \SQM\3\ \SQF\S\5\ \SQT\To accelerate a 50 g mass at the rate of 20 mm/s2 requires a force of ____ N.\ \SQH\F = ma\ \SQANS\.001\ \SQLAST\No, (0.05 kg) x (0.02 m/s2) = 0.001 N\ \SQA\T\CONTAINS\.001\Right. (0.05 kg) x (0.02 m/s2) = 0.001 N\ \5\ \SQT\To accelerate a 20 kg mass at the rate of 10 m/s2 requires a force of ____ N.\ \SQANS\200\ \SQH\F = ma\ \SQANS\200\ \SQLAST\No, (20 kg) x (10 m/s2) = 200 N\ \SQA\T\=\200\Right. (20 kg) x (10 m/s2) = 200 N\ \DQ\ \DCW\ \SQ\ \SQs at the rate of 2 m/s2 requires a force of ____ N.\ \SQH\F = ma\ \SQANS\4\ \SQLAST\No, (2 kg) x (2 m/s2) = 4 N\ \SQA\T\=\4\Right. (2 kg) x (2 m/s2) = 4 N\ \SQA\T\=\4.0\Right. (2 kg) x (2 m/s2) = 4 N\ \DQ\ \DCW\ \SQ\ \SQM\3\ \SQF\S\ requires a force of ____ N.\ \SQH\F = ma\ \SQANS\2\ \SQLAST\No, (2 kg) x (1 m/s2) = 2 N\ \SQA\T\=\2\Right. (2 kg) x (1 m/s2) = 2 N\ \SQA\T\=\2.0\Right. (2 kg) x (1 m/s2) = 2 N\ \DQ\ \DCW\ \SQ\ \SQF\S\5\ \SQT\To accelerate a 2 kg masTCNW\F = ma\ \SCH\1\1\ \DTN\ The force, F, required to accelerate an object of mass, m, at the rate a, is equal to the mass times the acceleration.\ \SQS\2\Y\ \DCW\ \SQ\ \SQM\3\ \SQF\S\5\ \SQT\To accelerate a 2 kg mass at the rate of 1 m/s2No, A = F/m = (3 N)/(3 kg) = 1 m/s2\ \SQA\T\=\c\Right. A = F/m = (3 N)/(3 kg) = 1 m/s2 of acceleration.\ \DQ\ \SQSEND\ \DCW\ \DTNNW\ The formula a = F/m, can be rearranged to give the more common form of Newton's Second Law:\ \SCH\2\3\ \DSQF\S\1\ \SQT\How rapidly will a 3 kg object accelerate when acted on by a force of 3 N? a. 0 m/s2 d. 3 m/s2 b. 0.3 m/s2 e. 9 m/s2 c. 1 m/s2 f. 27 m/s2\ \SQH\Remember, a = F/m\ \SQANS\c\ \SQLAST\ c. 1 m/s2 f. 50 m/s2\ \SQH\Remember, a = F/m\ \SQANS\b\ \SQLAST\No, a = F/m = (2 N)/(10 kg) = 0.2 m/s2\ \SQA\T\=\b\Right. a = F/m = (2 N)/(10 kg) = 0.2 m/s2\ \SQA\F\=\e\No, a = F/m, not m/F.\ \DQ\ \DCW\ \SQ\ \SQM\3\ \kg) = 0.5 m/s2\ \SQA\F\=\d\No, a = F/m, not m/F.\ \DQ\ \DCW\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\How rapidly will a 10 kg object accelerate when acted on by a force of 2 N? a. 0 m/s2 d. 2 m/s2 b. 0.2 m/s2 e. 5 m/s2 ce of 2 N? a. 0 m/s2 d. 2 m/s2 b. 0.5 m/s2 e. 5 m/s2 c. 1 m/s2 f. 80 m/s2\ \SQH\Remember, a = F/m\ \SQANS\b\ \SQLAST\No, A = F/m = (2 N)/(4 kg) = 0.5 m/s2\ \SQA\T\=\b\Right. A = F/m = (2 N)/(4 \SQANS\e\ \SQLAST\No, (5 N)/(1 kg) = 5 m/s2\ \SQA\T\=\e\Right. a = F/m = (5 N)/(1 kg) = 5 m/s2\ \SQA\F\=\b\No, a = F/m, not m/F.\ \DQ\ \DCW\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\How rapidly will a 4 kg object accelerate when acted on by a forDCW\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\How rapidly will a 1 kg object accelerate when acted on by a force of 5 N? a. 0 m/s2 d. 2 m/s2 b. 0.2 m/s2 e. 5 m/s2 c. 1 m/s2 f. 50 m/s2\ \SQH\Remember, a = F/m\\ \SQA\F\=\.5\No, m = F/a, not a/F\ \DQ\d \DCW\ \SQ\ \SQM\3\ \SQF\S\3\ \SQT\If the force of gravity accelerates a falling 1 kg mass at the rate of 9.8 m/s2, the force is ____ N.\ \SQH\F = ma\ \SQANS\9.8\ \SQLAST\No, F = ma = (1 kg) x (9.8 m/s2) = 9.8 N\ \SQA\T\=\9.8\Right. F = ma = (1 kg) x (9.8 m/s2) = 9.8 N\ \SQA\T\=\10\More accuracy, please. \ \DCW\ \SQ\ \SQM\3\ \SQF\S\3\ \SQT\If the force of gravity accelerates a falling 5 kg mass at the rate of 9.8 m/s2, the forcSFF\0\\ \DF\ADDV5\1\1\1\0\ \SWT\\\12\\ \DTN\ Two forces were applied to the ball at the same time. Forces applied at the same time are called concurrent forces To add them together, you must place the tail of one vector at the head of t\SWT\\\M\16\ \XERASE\84\55\91\89\ \XERASE\94\92\127\98\ \SA\ \DFA\BALL1\77\90\ \SAS\75\\1\\ \SAL\16\2\2\-2\-2\ \DA\ \XERASE\170\0\183\10\ \SWT\M\M\17\M\ \DCW\ \DTN\ The ball moves up and to the right. How could we have predicted this?\ \LOT\87\95\ \XDRAW\40\0\-3\-3\3\3\-3\3\ \XPLOT\87\95\ \XDRAW\0\-40\-3\3\3\-3\3\3\ \SWT\\\17\\ \DTN\ Here you see a ball with two equal forces being applied to it: one from the left and one from below.\ \DTNNW\ Let's see which way it moves.\ net force is applied. The object does not move at all.\ \DTN\ The net force can be calculated by adding up all the forces being applied. This is not as easy as it might sound, because each force is a vector.\ \DCW\ \DFB\B2\81\90\ \XPther\ \SQA\F\CONTAINS\left\No, the force isn't greater toward the left.\ \SQA\F\CONTAINS\right\No, the force isn't greater toward the right.\ \DQ\ \SFF\0\\ \DF\ADDV3\1\1\1\0\ \DTN\ Since two equal and opposite forces are being applied, no ch direction will it move - left, right, or neither?\ \SQH\Figure out the direction in which the force is greater.\ \SQANS\neither\ \SQLAST\No, the ball won't move either way.\ \SQA\T\=\neither\Right, it won't move either way.\ \SQA\T\CONTAINS\nei\SA\ \DFA\BALL1\245\10\ \SAS\75\\1\\ \SAL\14\-2\-2\\\ \SAL\1\-29\\\\ \DA\ \XERASE\9\10\22\20\ \GCR\ \SWS\1\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\7\ \SQT\If the same force is applied simultaneously to both the left and right sides of a ball, in whi0\ \SAS\75\\1\\ \SAL\16\2\2\\\ \DA\ \XERASE\243\10\256\20\ \GCR\ \SWT\\\6\\ \DCW\ \DTNNW\ When the same force is applied to the right side, the ball accelerates toward the left.\ \DFB\B2\245\10\ \XPLOT\250\15\ \XDRAW\-40\0\3\3\-3\-3\3\-3\ acting on the refrigerator simultaneously.\ \SWS\1\ \DCW\ \SWT\\\6\\ \DTNNW\ When a force is applied to the left side of this ball, it accelerates toward the right.\ \DFB\B2\0\10\ \XPLOT\5\15\ \XDRAW\40\0\-3\-3\3\3\-3\3\ \SA\ \DFA\BALL1\0\1 If you push gently against the side of a refrigerator, it probably won't move at all. Yet Newton's second law states that when a force is applied to an object, it accelerates.\ \DTN\ The problem here is that there are actually two forcesrn to Main Menu\ \DM\ \BMC\1\FRICTION ADD\ \BMC\2\FRICTION FRICTION\ \BMC\3\FRICTION RESOLVE\ \BMC\4\FRICTION INCLINE\ \BMC\5\FRICTION VOCAB\ \BMC\6\FRICTION REVIEW\ \BL\FORCE\ \L\FRICTION ADD\ \SCH\2\2\ \DTCNW\Adding Vectors\ \SCH\1\1\ \DTN\RVR\2\ \BESC\ \L\ACCEL REVIEW\ \RRQ\2\ \BESC\ \L\FRICTION\ \SM\ \SMT\Friction & Vectors\ \SMO\Adding Vectors\ \SMO\Friction\ \SMO\Resolving Vectors\ \SMO\Friction on an Incline\ \SMO\Vocabulary Review\ \SMO\Review Questions\ \SMO\Retu falling 10 kg mass at the rate of 9.8 m/s2, the force is ____ N.\ \SQANS\98\ \SQLAST\No, F = ma = (10 kg) x (9.8 m/s2) = 98 N\ \SQH\F = ma\ \SQA\T\=\98\Right. F = ma = (10 kg) x (9.8 m/s2) = 98 N\ \DQ\ \SQSEND\ \BESC\ \L\ACCEL VOCAB\ \e is ____ N.\ \SQH\F = ma\ \SQANS\49\ \SQLAST\No, F = ma = (5 kg) x (9.8 m/s2) = 49 N\ \SQA\T\=\49\Right. F = ma = (5 kg) x (9.8 m/s2) = 49 N\ \DQ\ \DCW\ \SQ\ \SQM\3\ \SQF\S\3\ \SQT\If the force of gravity accelerates ahe other.\ \DCW\ \DF\ADDV6\1\0\1\0\ \SWT\\\R2\\ \DTN\ Now, draw a new vector from the tail of the first one to the head of the second one.\ \DF\ADDV7\1\0\1\0\ \SWT\\\R2\\ \DCW\ \DTN\ This resultant vector is the sum of the two concurrent vectors. It has a length that indicates the magnitude of the resulting vector. It has a direction that indicates the direction of the resulting vector.\ \DCW\ \DTN\ The vector points in the direction the ball moves when pushed by both of\DQ\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\3\ \SQT\Two vectors of magnitudes 7 N and 10 N are applied to an object concurrently. The smallest possible magnitude for the resultant vector is ____ N.\ \SQANS\3\ \SQLAST\No, the smallest resultant vectortors are parallel and pointing in the same direction.\ \SQA\T\=\17\Right. The largest is 17 N (10 N + 7 N), obtained when the vectors are parallel and pointing in the same direction.\ \SQA\F\=\3\No, 3 N is actually the smallest force possible!\ The largest possible magnitude for the resultant vector is ____ N.\ \SQANS\17\ \SQH\The resultant vector will be greatest if they are pointing in the same direction.\ \SQANS\17\ \SQLAST\No, the largest is 17 N (10 N + 7 N), obtained when the vec8\0\ \XDEL\20\ \DFB\ADDV12D\48\0\ \DTN\ It can be as small as the difference between the two magnitudes.\ \SWS\1\ \DCW\ \SQS\3\Y\ \SQ\3\Y\ \SQM\3\ \SQF\S\3\ \SQT\Two vectors of magnitudes 7 N and 10 N are applied to an object concurrently. e added, both the magnitude and the direction of the resultant vector are dependent on the directions of the two vectors.\ \DFB\ADDV12B\48\0\ \DTN\ The magnitude can be as great as the sum of the two magnitudes.\ \DCW\ \XDEL\5\ \DFB\ADDV12C\4hose pushing toward the left.\ \SQA\T\CONTAINS\left\No! The total forces pushing toward the right are greater than those pushing toward the left.\ \DQ\ \SQSEND\ \SWS\1\ \DCW\ \SFF\0\\ \DFB\ADDV12\48\0\ \SWT\\\13\\ \DTN\ When two vectors arer (LEFT or RIGHT)?\ \SQANS\Right\ \SQLAST\No, more force is pushing toward the right (10 N) than toward the left (7 N), so the net force pushes to the right.\ \SQA\T\CONTAINS\right\Yes! The total forces pushing toward the right are greater than tthe right and a total of 7 N pushing to the left, so the result is 10 - 7 = 3 N.\ \DQ\ \DCW\ \SQ\3\Y\ \SQM\2\ \SQF\S\7\ \SQT\Is the force pushing the ball to the left or the right?\ \SQANS\right\ \SQH\In which direction is the total force great one way and subtract the forces going the other.\ \SQANS\3\ \SQLAST\No, there is a 10 N force pushing toward the right and a total of 7 N pushing to the left, so the result is 10 - 7 = 3 N.\ \SQA\T\=\3\Right. There is a 10 N force pushing toward des of the forces working in the opposite direction.\ \SQS\3\Y\ \SFF\0\\ \DF\ADDV11\1\0\1\0\ \SWT\\\R2\\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\3\ \SQT\The magnitude of the net force acting on this ball is ____ N.\ \SQANS\3\ \SQH\Add the forces goingt forces are applied, the direction of the resultant vector will be along the same line as the forces.\ \DTN\ The magnitude of the force is calculated by adding the magnitudes of all the forces working in one direction and subtracting the magnituesultant vector is drawn.\ \SWS\1\ \DCW\ \DFB\B2\4\10\ \XPLOT\9\15\ \XDRAW\30\30\-3\0\3\0\0\-3\ \SA\ \DFA\BALL1\0\10\ \SAS\75\\1\\ \SAL\16\2\2\2\2\ \DA\ \XERASE\159\166\172\176\ \DF\ADDV11\1\1\1\0\ \SWT\\\R2\\ \DTN\ When parallel concurrenXERASE\170\0\183\10\ \GCR\ \DF\ADDV8\1\1\1\0\ \SWT\\\12\\ \DTN\ Here are four concurrent forces. \ \DF\ADDV9\1\1\1\0\ \SWT\\\12\\ \DTN\ Here they have been combined, head to tail.\ \DCW\ \DF\ADDV10\1\0\1\0\ \SWT\\\12\\ \DTN\ Now the r these forces.\ \SWS\1\ \DCW\ \SWT\M\M\17\M\ \DTNNW\ Any number of concurrent forces can be combined in this manner:\ \DFB\B2\81\90\ \XPLOT\87\95\ \XDRAW\30\-30\-3\0\3\0\0\3\ \SA\ \DFA\BALL1\77\90\ \SAS\75\\1\\ \SAL\16\2\2\-2\-2\ \DA\ \ is seen when they are parallel but pointing in opposite directions: 10 N - 7 N = 3 N\ \SQH\The resultant vector will be smallest if they are parallel but pointing in opposite directions.\ \SQA\F\CONTAINS\-\No, the magnitude of a vector can never be less than zero!\ \SQA\T\=\3\Right. With the two forces parallel but pointing in opposite directions, the magnitude is 10 N - 7 N = 3 N.\ \DQ\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\3\ \SQT\Three vectors of magnitudes 4 N, 5 N, and 10 N are applied tMon Aug 13 14:47:00 1990 \L\acceleration\ \2\the rate of change of velocity.\ \L\average speed\ \1\the total displacement (the straight-line distance from the starting point to the final position) divided by elapsed time.\ \L\average acceleration\ nature of the two surfaces (how "slippery" they are) and the magnitude of the force holding the object to the surface (normally its weight).\ \DCW\ \DTNNW\ In general, the starting friction, F, is proportional to the force holding the object way. When the applied force becomes great enough, the object will begin to move. The applied force is now greater than the starting friction.\ \SWS\1\ \DCW\ \DF\FRICTN2\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ This starting friction depends on theen the two objects and opposite to the direction of the applied force.\ \DTN\ The magnitude of the frictional force is equal to the applied force (Fa).\ \DCW\ \DTN\ There is a limit to the amount of frictional force that can develop this , which opposes the movement.\ \DTN\ Like all forces, frictional force has both a magnitude and a direction.\ \DTN\ For an object at rest with a force applied to one side, the direction of the frictional force is parallel to the surface betwehen one is sliding over the other, or when a force is acting which tends to slide one over the other.\ \DTN\ When an object is at rest on a surface and a small force (Fa) is applied to one side of the object, a frictional force (Ff) developsave disappeared.\ \DTN\ Very strange!\ \DCW\ \SFF\0\\ \DF\FRICTN1\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ This mysterious force is friction.\ \DTN\ Friction is a force that acts to resist movement. It acts at the surface between two objects, w Therefore, there must be an equal and opposite force acting on it simultaneously, so that the resultant force is zero.\ \DTN\ When you stop pushing, the refrigerator continues to sit there. Therefore, that equal and opposite force must hther: F = (10 N) - (4 N) - (5 N) = 1 N\ \DQ\ \SQSEND\ \BESC\ \L\FRICTION FRICTION\ \SCH\2\2\ \DTCNW\Friction\ \SCH\1\1\ \DTN\ When you push on the side of a refrigerator, you are applying a force to it. But it doesn't move.\ \DTN\ ng the two smaller vectors in one direction and the largest one in the other: F = (10 N) - (4 N) - (5 N) = 1 N\ \SQA\T\=\1\Right. The smallest result is obtained by pointing the two smaller vectors in one direction and the largest one in the opossible magnitude for the resultant vector is ____ N.\ \SQH\The smallest result is obtained by pointing the two smaller vectors in one direction and the largest one in the other.\ \SQANS\1\ \SQLAST\No, the smallest result is obtained by pointi result is 19 N, obtained when all three vectors point in the same direction; 10 N + 5 N + 4 N = 19 N.\ \DQ\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\3\ \SQT\Three vectors of magnitudes 4 N, 5 N, and 10 N are applied to an object concurrently. The smallest ude of the resultant vector is equal to the sums of the individual magnitudes.\ \SQANS\19\ \SQLAST\No, the largest result is 19 N, obtained when all three vectors point in the same direction; 10 N + 5 N + 4 N = 19 N.\ \SQA\T\=\19\Right. The largesto an object concurrently. The largest possible magnitude for the resultant vector is ___ N.\ \SQANS\19\ \SQH\The largest result will be obtained if they all point in the same direction.\ \SQH\If they all point in the same direction, the magnit \2\the total change of velocity divided by the elapsed time.\ \L\coefficient of starting friction\ \3\the ratio of the force required to start sliding motion to the normal (perpendicular) force between the two surfaces.\ \L\concurrent forces\ \3to the surface (F):\ \SCH\2\3\ \DTC\F = F\ \SCH\1\1\ \DTN\ The proportionality factor, , (the Greek letter, mu) is called the coefficient of starting friction. It is a function of the two surfaces between which sliding might occur.\-motion\Newton's first law of motionnewtonsecond%slopespeed starting friction tangent time' unit vector velocity +accelerationdaverage speedaverage accelerationhcoefficient of starting frictionconcurrent forces;directiondistanceforceinstantaneous velocityXkilogrammagnitudemassfmetermilliular direction.\ exists or something occurs.\ \L\unit\ \1\the basic measure of a property, such as 1 meter, 1 sec, or meter per second per second.\ \L\vector\ \1\a measurement that includes a direction as well as a magnitude.\ \L\velocity\ \1\a speed in a particing friction\ \3\the maximum frictional force resisting motion between two objects in contact with one another.\ \L\tangent\ \1\a line drawn so as to touch a curve at only a single point.\ \L\time\ \1\a measured period during which a condition ne sixtieth of a minute.\ \L\slope\ \1\an indication of how steep a line is - the ratio of the vertical displacement to the horizontal displacement along the line.\ \L\speed\ \1\the rate of movement, measured as distance per unit time.\ \L\startton\ \2\the basic unit of force - the force needed to accelerate a 1 kilogram mass at a rate of 1 meter per second per second.\ \3\a vector obtained through the addition or subtraction of other vectors.\ \L\second\ \1\the basic unit of time - oin relation to objects that are considered stationary.\ \L\Newton's first law of motion\ \2\an object at rest will remain at rest and an object in motion will remain in motion with constant velocity unless an unbalanced force acts upon it.\ \L\newthe amount of material in an object, defined by its inherent resistance to acceleration.\ \L\meter\ \1\the basic metric measure of length (39.37 inches).\ \L\milli-\ \1\a prefix meaning "thousandth."\ \L\motion\ \1\the displacement of an object which it is measured approaches zero.\ \1\a prefix meaning "thousand."\ \L\kilogram\ \2\the basic metric unit of mass, equal to 1,000 grams (2.2046 pounds).\ \L\magnitude\ \13\a numerical measure of size or quantity.\ \L\mass\ \2\a measure of t resists the relative motion of objects in contact with one another.\ \1\the magnitude component (size) of the instantaneous velocity vector.\ \L\instantaneous velocity\ \1\the limit which the average velocity approaches as the time interval over \forces simultaneously acting on a body.\ \L\direction\ \1\the line or course along which an object moves.\ \L\distance\ \1\the length of a line segment between two points.\ \L\force\ \2\a push or pull exerted on an object.\ \3\the force tha \SCH\2\3\ \DCW\ \DTNNW\ F \ \DTNW\ = -- \ \DT\ F\ \SCH\1\1\ \DTN\ Notice that is the ratio of two forces. As a result, it has no units.\ \DTN\ If the starting force for a 5 kg wooden cube sitting on a plank of wood is 24.5 N, = (24.5 N)/(5 kg)(9.8 m/s2) = (24.5 N)/(49.0 N) = 0.5 The coefficient of starting friction is one-half.\ \SWS\1\ \DCW\ \SCH\1\3\ \DTNNW\ F = -- œKiKLiL( jQ8PMmJŕ MKL ` Lj` jMmPiQ+KL lmN Ft j敥` & & &mbce`MR **&j =k T ȱ=k T` U+MP*RjCSNOPQ 9 )`J `iJ8`J hU(T`U(T``h(`(`^` ȱ8 !!ȱ"%ȱ## z` `)S`be8h` Hy6x27xɠ?2 ex7x2`6x Ny` ,` Vx`)` 0j)$202 x?2Lx%2 Ny``2``ȱ^e"%L z%``%`` x x y x` x x8% x%H zh y xL zͥ6ک6 % 4w= Ϡȱi故 FteĒƓ摥e故Lio`R`R` P׶oPطoϥS*@S Po@, Po@- W) Wo ) WR( Ro(o(W'R,)mH%hI1ĔLMnŖ2m EP *m,+m +m,,m ,m,-m -mLmƕ晥e摅Lm`hires.tbldrnȱesnpn fȩnȭrnȭsnȑȑȩ rtpn T nRWP ) ,SQĔƕ春e摅Ll` 0@`U**UȱȱȱȱƠȱi摅.m2me Fte)HJhkŗ].mR E,&m &m,'m 'm,(m (m,)mLmHĜhLmEJh )mh9 r B ,x3{ h){ h2{ x r@@*x+N{ H{ R{ * r+ <{ r c{ r g{ xx*xr+֕{ ؏{ r r + * r+ t{ h{ hxط{@x{x+{ r+ r׾{ r@ @+ { +{=;<{IO error: $ 3aȱbȱiȱȱȱmm|i} 轠ʥ|}ʊJȱ|Șe||}i}ĩ`fjkcdn{p|8}|魁LXLBW% qqW$" X`XL ȱ Lъ%" "LL%"#ee%" H H "%"" L LLB L QEXQgRjTWYŠX` ȱeeLވW8iiLJH H WL`  ȱLo  ȱL) LH H W`XL =ȱLX`XL qqWX`!(p00LqHH V ȱ `  0  )` LO恥 L iȱiHH LI ȱ ` ȱXL  ȱLK)LKWł僰L % L!HHH/H <L]H H &' LuLm&m' Lm&m'ȑHHHHHH 8& &''m&m'HHHHqHH )LuLB&'LB&'LuLbHH L!m&&m''&'Lm&&m'')L!L솊HHH(H <L!HHH+H <'i')Lu) LG LG m&m' )L…`Liƅi`Lc)L Em&m')@LLm&&m''LZ  @ A© é~+=>= W^&׀D© өé ĩƩ~+ L Ϥȩ @ +  -ѳc8 =<*~ @%ft(f  Bг SfR)v&  6<钐u s9 rӋ:1 ӳS% ӏ+ @ W^&=8 9΁'޲<޲3W D F F* `Negohi f8hijk f n QRTVlHQp h $ab Liةɩ뭖hiL}4fg LՋ%p0L8%%" `!$$p00L#ee%p00LH H L%$" W`W$" `X WLW%$" L Q  , LQXL8W% qqW{" X`XL ȱ L%"HH "LL%"HH "LHH LLH H L QENQbReTyWYX` ȱ L6%p0L8%%" ` ȑLⓥ  0ȘPILⓥ 0ȘPILⓥ ȄLⓥ  ȄLⓥ  ȄLⓥ ȄLⓥ ȄL8 LⓥI I Lⓥ  L Lⓥ L& L : םL : L : Lⓠ   8  8 `e  e L8  Lⓦ  L&   L8f f  LF f%  % Lⓥ   LⓥE  E Lⓥ  ȄLⓥ  ȄLⓥ ȘPILⓥ  ȘPI & e eȱL⓱ȱ L⓱ L V  L V Lⓥ L VL⓱ ȱ i`J e e fff fL L LLⓠ& &&8  `  LL  Lⓥ E L⓱  eL :eeL :Lⓥ :Lⓥ eLⓥ Lⓥ eLⓥ Lⓥ ȱ i ʱȱ iȱȱLeȱe ȥȥ ȰL g P :ee L :e e L  L :  L : L :L ee L : g L P g L P gL P gL P L P ҖL L̖ P ҖL8ȥ i`l :   ȥ Lⓩ*&" eeȱL  L) LⓄ LⓊ) LⓄ LⓊ)L)e L g !ȱ L P ! ȱ L g P !ȥ L g P !ȱ*&" eeȱ Lⓠ   ȱ Lⓩ*&" ee ȱ Lⓠ    ȱ Lⓩ*&" eeȥ Llhh hihi8  8   qȥqi HvHl    lhh hh8   qȥ qi il8iehHHl ȱ `hhȱ@i@ʱȱ iȱȱ lhh8iehȱ Hȥq` `e e `ȥ 2 e e  e e  `  `   e  `   ` Ȅ`  `hh HȱhlȱȱHȱ h` P X Ҧf* ӌڌC >@Ϥ Q RϠ © `   `A[i  A[i ʰ۠ Ȅ`     ߏL`l K 9786 H66Аl86`Ϣ   )   `Ϣ  )  ) a{iઘ)* &  e 8 ߏ `)  ߏ  i   )0" H $(*0ޠ L" A©@é~v ϮϢ  )) )c )LK8LL $ȥ ȥ Mȩȅ  ) ȱ  )Li i e WLQW{" `W ȑLQW ȑLQ Q Ɍ &Q ȱ LmWWiLX`  ȱp00LWL Q_ou HH ` ) )Ƀȑ `#ʊH $(*)+!hʙlLⓢ`ee`ȱHih`ȱi`eȱei`8ȊL ML "&KGC?Y]aeimquy} #'+/37;?Cfb^Z{qqqqqqqqqqqqqqq *49CHNTc hytׅ'/IPkN,6@L} Tbnv  e. 9.3 b. 0.108 d. 1.054\ \SQH\Remember, in this case F = ma = 9.8 x 1 = 9.8 N.\ \SQH\Remember, = F/F\ \SQANS\c\ \SQLAST\No, = F/F and F = ma = 9.8 x 1 9.8 N, so = (9.3 N)/(9.8 N) = 0.949\ \SQA\T\=\c\Right, = F/F and F 6789:;<=>?pabcdefghijklmnopqrstuvwxyz{|mn   DEV_INFO04/86CON:@PR:SER: 0@P`p@ allocatingsci.fnt( press a key to reboot'*U*UU*U*"D"Dn];w];wn @A'ԩ/...@{| ~`} ^ !"#$%&'()*+,-./012345to be? 1. 4 m/s (Mars) 2. 10 m/s (Earth) 3. 20 m/s (2 x Earth) 4. 30 m/s (Jupiter) YyNn123451234pdisk1.sfgError reading figure file Error allocating main_art_buf01.02.0kmerror m/sWhat horizontal velocity do you want to give your ball? 1. 10 m/s 2. 50 m/s 3. 80 m/s 4. 100 m/s 4 m/s10 m/s20 m/s30 m/sǤHow great do you want the gravitational acceleration ? Press Y or N. u]2$*2200 m500 m1,000 m1,500 m"&*.26:>#'+/37;?#'+/37;? $(,048< $(,048"&*.26:>#'+/37;?#'+/37;?((((((((((((((((((((((((((((((((PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP $(,048< $(,048"&*.26:>#'+/37;?#'+/37;?equired to start it sliding? a. 0.04 g d. 0.04 kg b. 0.25 g e. 0.25 kg c. 2.50 g f. 2.50 kg\ \SQH\Remember that F = F/ = 0.1/0.04 = 2.5 N\ \SQH\Remember, since F = ma, the mass = F/9.8\ \SQH\mass = 2.5x 9.8 = 39.2, so F = 0.74 x 39.2 = 29 N\ \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\1\ \SQT\ The coefficient of starting friction for a Teflon coated book on a Teflon coated surface is 0.04. What is the mass of this book if 0.1 N of force is r2.96 N e. 29.0 N c. 4.00 N f. 39.2 N\ \SQH\Remember that F = 4 x 9.8 = 39.2 N\ \SQH\Remember, F = F\ \SQANS\e\ \SQLAST\No, F = F and F = 4 x 9.8 = 39.2, so F = 0.74 x 39.2 = 29 N\ \SQA\T\=\e\Right. F = F and F = 4 n newtons.\ \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\1\ \SQT\What is the force required to start a 4 kg block of steel sliding along a sheet of steel, if the coefficient of starting friction is 0.74? a. 0.74 N d. 5.41 N b. s the gravitational acceleration constant, 9.8 m/s2: 3.5 x 9.8 = 34.3 N\ \SQA\T\=\f\Right. The force is the mass times the gravitational acceleration constant, 9.8 m/s2: 3.5 x 9.8 = 34.3 N\ \SQA\F\IS CONTAINED IN\abc\No, the force must be i\DTN\ Here is a simple vector.\ \DTN\ Any two line segments that connect the two ends of this vector could stand for vectors which add up to this one.\ \XCOLOR\\\1\ \XPLOT\46\6\ \XDRAW\119\4\-4\-4\ \XPLOT\165\10\ \XDRAW\-4\4\ \XPLOT\82\52\ \XDRAW\-36\-46\0\4\ \XPLOT\46\6\ \XDRAW\4\0\ \XDEL\10\ \XPLOT\82\52\ \XDRAW\83\0\-4\-4\ \XPLOT\165\52\ \XDRAW\-4\4\ \XPLOT\165\54\ \XDRAW\0\-44\-4\4\ \XPLOT\165\10\ \XDRAW\4\4\ \XDEL\10\ \XPLOT\82\52\ \XDRAW\126\8\-4\-4\ \XPLOT\208\60\ \XDRA \SQ\3\Y\ \SQM\3\ \SQF\N\5\\QH\ \SQT\A 10 kg block rests on a frictionless surface raised 30 above the horizontal. The force accelerating the block down the plane is ____ N.  sin(30) = 0.500 cos(30) = 0.866 tan(30) = 0.577\ \SQANS\49 N\ \SQH 5 m/s2.\ \SQA\T\=\d\Right. a = F/m = (20 N)/(4 kg) = 5 m/s2.\ \SQA\F\=\f\No, 20 N is the force, not the acceleration.\ \SQA\F\=\a\No, a = F/m, not m/F.\ \DQ\ \BQESC\ \SQSEND\ \SWS\1\ \SQS\3\Y\ \SFF\0\\ \DF\INCLTEST\1\1\1\0\ \SWT\\\7\\will this 4 kg block accelerate down the incline? a. 0.2 m/s2 c. 2 m/s2 e. 10 m/s2 b. 1.0 m/s2 d. 5 m/s2 f. 20 m/s2\ \SQH\Remember, a = F/m.\ \SQH\Remember, F = 20 N and m = 4 kg.\ \SQANS\d\ \SQLAST\No, a = F/m = (20 N)/(4 kg) =erating force is the force parallel to the plane, shown here to be 20 N.\ \SQA\T\CONTAINS\20.\Right. The accelerating force is the force parallel to the plane, shown here to be 20 N.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\How rapidly ock down the plane is ____ N.\ \SQH\Remember, the accelerating force is the force parallel to the plane, F.\ \SQANS\20\ \SQLAST\No, the accelerating force is the force parallel to the plane, shown here to be 20 N.\ \SQA\T\=\20\Right. The accel = 39.2 N and = 30:  F = F sin() = 39.2 x .500 = 19.6 N  F = F cos() = 39.2 x .866 = 33.9 N\ \SQS\3\Y\ \SFF\0\\ \DF\RESOLVED\1\1\1\0\ \SWT\\\R2\\1\ \SQ\ \SQM\3\ \SQF\S\5\ \SQT\The magnitude of the force acting to accelerate the bl calculate the magnitude of F and F by trigonometry.\ \DTN\ If the angle of the plane with the horizontal is represented by the Greek letter theta (), then: \ \SCH\2\3\ \DTCNW\F = F sin()\ \DTC\F = F cos()\ \SCH\1\1\ \DTN\ Since F. "Normal" is another word for "perpendicular."\ \XPLOT\63\84\ \XDRAW\6\2\ \XTEXT\\65\75\ \XPLOT\212\89\ \XDRAW\-3\15\ \XTEXT\\214\94\ \XTEXT\=Fg sin()\108\20\ \XTEXT\=Fg cos()\91\80\ \XPLOT\14\98\ \XDRAW\5\0\0\5\ \DCW\ \DTN\ We can \DCW\ \DTN\ The gravitational force, F, must be resolved into one component parallel to the surface (F) and one component perpendicular to the surface (F).\ \DTN\ The term normal force is used for the force perpendicular to the surfaceational force which is parallel to the direction of movement contributes to the object's acceleration.\ \XPLOT\63\24\ \XDRAW\40\13\-2\-4\ \XPLOT\103\37\ \XDRAW\-4\1\ \XPLOT\103\37\ \XDRAW\-40\59\ \XTEXT\F\77\80\ \XTEXT\N\84\84\ \XTEXT\Fp\94\20\ \XPLOT\63\24\ \XDRAW\0\72\-4\-4\ \XPLOT\63\96\ \XDRAW\4\-4\ \XTEXT\Fg\42\60\ \DCW\ \DTN\ The force of gravity, F, is 4 x 9.8 = 39.2 N. But this force is not directed down the inclined plane.\ \DTN\ Only that component of the gravitowed to slide down a frictionless inclined plane, tilted 30 from the horizontal. How rapidly will it accelerate down this plane?\ \DTN\ To calculate this rate, you must determine the force pushing the block down the plane.\ \XCOLOR\\\1\ ew of these pairs.\ \DTN\ The process of breaking one force into two or more forces which are equivalent to the first, is called resolving a vector into two or more components.\ \DF\RESOLV3\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ A 4 kg block is allW\-4\4\ \XPLOT\210\60\ \XDRAW\-45\-50\0\4\ \XPLOT\165\10\ \XDRAW\4\0\ \XDEL\10\ \XPLOT\82\52\ \XDRAW\-40\6\4\-4\ \XPLOT\42\58\ \XDRAW\4\4\ \XPLOT\42\58\ \XDRAW\123\-48\-4\0\ \XPLOT\165\10\ \XDRAW\0\4\ \XCOLOR\\\3\ \DTN\ These are just a f\Remember, the total force is (10 kg) x (9.8 m/s2)\ \SQH\Remember, the accelerating force is the component of the total force parallel to the plane.\ \SQANS\49\ \SQLAST\No, the force is F sin(), where F = 98 N, so F = 0.5 x 98 N = 49 N.\ \SQA\T\=\49\Right. The force is F sin(), where F = 98 N, so F = 0.5 x 98 N = 49 N.\ \SQA\F\=\5\No, the mass is 10 kg, but the total force is (10 kg) x (9.8 m/s2) = 98 N.\ \SQA\F\=\50\More accuracy, please.\ \DQ\ \SQSEND\ \BQESC\ \DCW\ \SQ\3\N\e INCREASE, DECREASE, or remain the SAME?\ \SQH\Does the parallel component increase or decrease compared to the total force?\ \SQANS\increase\ \SQLAST\No, the rate of acceleration increases. The parallel force component increases while the massroportion to the object's weight, so a = F/m remains constant.\ \SQA\T\CONTAINS\same\ \DQ\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\10\ \SQT\When the angle of a frictionless inclined plane increases, does the rate of acceleration of a block on the planceleration INCREASE, DECREASE, or remain the SAME.\ \SQH\Does the parallel component increase or decrease compared to the total force?\ \SQANS\same\ \SQLAST\No, the rate of acceleration stays the same. The parallel force component increases in p F = F sin() and F = mg, so a = 9.65 m/s/s\ \SQA\F\=\a\No, the force is mg, not m.\ \DQ\ \SQSEND\ \SWS\1\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\7\ \SQT\When the mass of an object on a frictionless inclined plane increases, does its rate of ac2 b. 1.67 m/s2 e. 19.30 m/s2 c. 1.69 m/s2 f. 57.65 m/s2\ \SQH\F = mg\ \SQH\F = F sin()\ \SQH\a = F/m\ \SQANS\d\ \SQLAST\No, a = F/m, F = F sin() and F = mg, so a = 9.65 m/s/s\ \SQA\T\=\d\Right. a = F/m,F/sin().\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\4\ \SQF\S\1\\QH\ \SQT\Imagine a 10 kg block on a frictionless inclined plane raised 80 above the horizontal. How rapidly will the block accelerate down the incline? a. 0.96 m/s2 d. 9.65 m/sSQH\a = F/m\ \SQANS\b\ \SQLAST\No, a = F/m, F = F sin() and F = mg, so a = 1.67 m/s/s\ \SQA\T\=\b\Right. a = F/m, F = F sin() and F = mg, so a = 1.67 m/s/s\ \SQA\F\=\a\No, the force is mg, not m.\ \SQA\F\=\f\No, F = F sin(), not aised 10 above the horizontal. How rapidly will the block accelerate down the incline? a. 0.17 m/s2 d. 9.65 m/s2 b. 1.67 m/s2 e. 19.30 m/s2 c. 1.69 m/s2 f. 57.65 m/s2\ \SQH\F = mg\ \SQH\F = F sin()\ \o, a = F/m, F = F sin() and F = mg, so a = 4.9 m/s2\ \SQA\T\=\4.9\Right. a = F/m, F = F sin() and F = mg, so a = 4.9 m/s2\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\4\ \SQF\S\1\\QH\ \SQT\Imagine a 10 kg block on a frictionless inclined plane r\SQM\4\ \SQF\S\3\\QH\ \SQT\Imagine a 40 kg block on a frictionless inclined plane raised 30 above the horizontal. The block will accelerate down the incline at a rate of ____ m/s2.\ \SQH\F = F sin()\ \SQH\a = F/m\ \SQANS\4.9\ \SQLAST\N2.\ \SQH\F = mg\ \SQH\F = F sin()\ \SQH\a = F/m\ \SQANS\4.9\ \SQLAST\No, a = F/m, F = F sin() and F = mg, so a = 4.9 m/s2\ \SQA\T\=\4.9\Right. a = F/m, F = F sin() and F = mg, so a = 4.9 m/s2\ \DQ\ \BQESC\ \DCW\ \SQ\ 6 0.577 80 0.985 0.170 5.80\ \SCH\\1\ \SWT\\\7\\ \SQ\ \SQM\4\ \SQF\S\3\\QH\ \SQT\Imagine a 20 kg block on a frictionless inclined plane raised 30 above the horizontal. The block will accelerate down the incline at a rate of ____ m/s = F/m = (49 N)/(10 kg) = 4.9 m/s2\ \SQA\F\=\5\More accuracy, please.\ \DQ\ \BQESC\ \SWS\1\ \SQS\3\Y\ \DCW\ \SCH\\3\ \DTNNW\ angle sin cos tan ----- ----- ----- ----- 10 0.170 0.985 0.172 30 0.500 0.86 \SQM\3\ \SQF\S\3\\QH\ \SQT\The block will accelerate down the incline at a rate of ____ m/s2.\ \SQH\a = F/m\ \SQH\F = 49 N.\ \SQH\a = (49 N)/(10 kg)\ \SQANS\4.9\ \SQLAST\No, a = F/m = (49 N)/(10 kg) = 4.9 m/s2\ \SQA\T\=\4.9\Right. a remains constant, so a = F/m increases.\ \SQA\T\CONTAINS\increase\Right. The parallel force component increases while the mass remains constant, so a = F/m increases.\ \DQ\ \BESC\ \L\FRICTION INCLINE\ \SCH\2\2\ \DTCNW\Friction on an Incline\ \SCH\1\1\ \DTN\ When a block is placed on an inclined plane, it may not slide at all. This will depend on the slope of the plane and the coefficient of starting friction.\ \DF\FINCLIN1\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ The maximal frictional e. 84.9 N c. 9.80 N f. 98.0 N\ \SQANS\d\ \SQLAST\No, the force is F sin(), where F = 98 N, so F = 0.5 x 98 N = 49 N.\ \SQH\Remember, the total force is (10 kg) x (9.8 m/s2)\ \SQH\Remember, the accelerating force is the compon`p~0p~`{`]@s~ x~@{?@}?@{?owNx `x?||nvn<\{8>` ?`pp8X8p}pn`y? |@@`}`~`}@w@;g| p|~?~?w{w^?n=p @pxx\l\x~8wp|@ ~@`0 kg block on an inclined plane raised 30 above the horizontal.\ \SWT\\\R2\\1\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\What is the magnitude of the force acting to accelerate the block down the plane? a. 5.00 N d. 49.0 N b. 8.66 N No, the maximal frictional force is greater than the force of acceleration, so it won't slide.\ \DQ\ \SQSEND\ \BQESC\ \SWS\1\ \DCW\ \SQS\3\Y\ \SCH\\3\ \DTNNW\ sin(30) = 0.500 cos(30) = 0.866 tan(30) = 0.577\ \SCH\\1\ \DTNNW\Imagine a 1o\ \SQLAST\No, the maximal frictional force is greater than the force of acceleration, so it won't slide.\ \SQA\T\CONTAINS\no\Right. The maximal frictional force is greater than the force of acceleration, so it won't slide.\ \SQA\T\CONTAINS\yes\If the maximal frictional force retarding a 4 kg block from sliding is 22 N and the force downhill parallel to the plane's surface is 20 N, will the block slide?\ \SQH\Is the force of acceleration larger than the maximal frictional force?\ \SQANS\Nis F sin(), where F = 39.2 N, so F = 0.5 x 39.2 N = 19.6 N.\ \SQA\F\=\20\More accuracy, please.\ \SQA\F\=\2\No, the mass is 4 kg, but the total force is (4 kg) x (9.8 m/s2) = 39.2 N.\ \DQ\ \DCW\ \SQS\3\Y\ \SQ\3\Y\ \SQM\3\ \SQF\S\3\ \SQT\e plane, F sin().\ \SQANS\19.6\ \SQLAST\No, the force is F sin(), where F = 39.2 N, so F = 0.5 x 39.2 N = 20 N.\ \SQA\T\=\19.6\Right. The force is F sin(), where F = 39.2 N, so F = 0.5 x 39.2 N = 19.6 N.\ \SQA\T\=\19.60\Right. The force izontal, the force accelerating the block down the plane will have a magnitude of ____ N.\ \SQANS\49\ \SQH\Remember, the total force is (4 kg) x (9.8 m/s2)\ \SQH\Remember, the accelerating force is the component of the total force parallel to thiction force then will be F = 0.65 x 34 = 22 N.\ \SWS\1\ \DCW\ \DTNNW\ (sin(30) = 0.500 cos(30) = 0.866 tan(30) = 0.577)\ \SWT\\\R2\\1\ \SQ\ \SQM\3\ \SQF\S\5\ \SQT\If a 4 kg block is placed on an inclined plane raised 30 above the hor\ \XERASE\151\97\172\107\ \XTEXT\34N\152\99\ \DCW\ \DTN\ Since F = F cos() and cos(30) = 0.866, the normal force holding the block to the plane is 39 x 0.866, or 34 N.\ \XERASE\70\12\92\23\ \XTEXT\22N\72\16\ \DCW\ \DTN\ The maximal fr that the block weighs 4 kg, that the angle of the plane is 30, and that the coefficient of starting friction is 0.65.\ \XERASE\116\72\137\81\ \XTEXT\39N\117\74\ \DCW\ \DTN\ Since F = ma, the gravitational force on the block is 4 x 9.8 = 39 N. force which can act to prevent the block from sliding is equal to the coefficient of starting friction, , times the normal force, F, holding the block to the plane:\ \DCW\ \SCH\2\3\ \DTC\F = F\ \SCH\1\1\ \DTN\ In this example, assumeent of the total force parallel to the inclined plane, F sin().\ \SQA\T\=\d\Right. The force is F sin(), where F = 98 N, so F = 0.5 x 98 N = 49 N.\ \SQA\F\=\a\No, the mass is 10 kg, but the total force is (10 kg) x (9.8 m/s2) = 98 N.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\What is the magnitude of the force holding the block against the plane? a. 5.00 N d. 49.0 N b. 8.66 N e. 84.9 N c. 9.80 N f. 98.0 N\ \SQANS\e\ \SQLAST\No, the fDQ\ \SQSEND\ \BESC\ \L\FRICTION VOCAB\ \RVR\3\ \BESC\ \L\FRICTION REVIEW\ \RRQ\3\ \BESC\ \L\EXPERIMENT\ \SCH\2\2\ \DTCNW\Falling Bodies\ \SCH\1\1\ \SWT\\\R2\\1\ \SQ\4\N\ \SQF\C\1\ \SCT\1\10\ \SQT\Do you want to see instructions for this cos()\ \SQANS\b\ \SQLAST\No, the forces are balanced when F = F cos(),so = F/[F cos()] = 49 / (98 x .866) = 0.58\ \SQA\T\=\b\Right. The forces are balanced when F = F cos(), so = F/[F cos()] = 49 / (98 x .866) = 0.58\ \hat coefficient of starting friction does the maximal frictional force just balance the accelerating force? a. 0.50 d. 1.33 b. 0.58 e. 1.73 c. 0.75 f. 2.00\ \SQH\F = F cos()\ \SQH\At equilibrium, F = F force is now greater than the force of acceleration.\ \DQ\ \SQSEND\ \BQESC\ \SWS\1\ \DCW\ \SQS\3\Y\ \SCH\\3\ \DTNNW\ F = F sin() = 49.0 N F = F cos() = 84.9 N\ \SCH\\1\ \SWT\\\R2\\1\ \SQ\ \SQM\4\ \SQF\S\1\ \SQT\For wNo it won't! The maximal frictional force is now greater than the force of acceleration.\ \SQA\T\CONTAINS\y\No it won't! The maximal frictional force is now greater than the force of acceleration.\ \SQA\T\CONTAINS\n\Right! The maximal frictionalQ\3\Y\ \SQM\3\ \SQF\S\7\ \SQT\In this case, with the maximal frictional force of 64 N and a force of acceleration of 49 N, will the block slide?\ \SQH\Is the force of acceleration greater than the maximal frictional force?\ \SQANS\no\ \SQLAST\ F = F cos()\ \SQANS\d\ \SQLAST\No, the maximal force, F is F cos() = 0.75 x 98 x .866 = 63.7 N\ \SQA\T\=\d\Right. The maximal force, F is F cos() = 0.75 x 98 x .866 = 63.7 N\ \DQ\ \BQESC\ \DCW\ \S \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\1\ \SQT\What will the maximum frictional force be if the coefficient of friction is 0.75? a. 5.0 N d. 63.7 N b. 42.4 N e. 84.9 N c. 49.0 N f. 98.0 N\ \SQH\F =l frictional force of 42 N.\ \SQA\T\CONTAINS\y\Right! The force of acceleration is greater than the maximal frictional force of 42 N.\ \SQA\T\CONTAINS\n\Yes it will! The force of acceleration is greater than the maximal frictional force of 42 N.\ictional force of 42 N and a force of acceleration of 49 N, will the block slide?\ \SQH\Is the force of acceleration greater than the maximal frictional force?\ \SQANS\yes\ \SQLAST\Yes it will! The force of acceleration is greater than the maxima maximal force, F is F cos() = 0.5 x 98 x .866 = 42.4 N\ \SQA\T\=\c\Right. The maximal force, F is F cos() = 0.5 x 98 x .866 = 42.4 N\ \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQM\3\ \SQF\S\7\ \SQT\With a maximal frf starting friction is 0.50, what will the maximum frictional force restraining sliding be? a. 5.0 N d. 49.0 N b. 8.7 N e. 84.9 N c. 42.4 N f. 98.0 N\ \SQH\F = F = F cos()\ \SQANS\c\ \SQLAST\No, theA\T\=\e\Right. The force is F cos(), where F = 98 N, so F = 0.866 x 98 N = 84.9 N.\ \SQA\F\=\b\No, the mass is 10 kg, but the total force is (10 kg) x (9.8 m/s2) = 98 N.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQM\3\ \SQF\S\1\ \SQT\If the coefficient oorce is F cos(), where F = 98 N, so F = 0.866 x 98 N = 84.9 N.\ \SQH\Remember, the total force is (10 kg) x (9.8 m/s2)\ \SQH\Remember, the accelerating force is the component of the total force parallel to the inclined plane, F cos().\ \SQ experiment?\ \SQH\Press Y or N\ \SQLAST\\ \SQA\T\IS CONTAINED IN\yn\\ \DQ\ \SWS\1\ \BIFESC\ \BIF\S\0\=\"n"\EXPBAL\ \DCW\ \DTN\ This simulation looks at the behavior of falling bodies. We are assuming that there is no frictional resistance to the object as it falls.\ \DTN\ When an object is thrown parallel to level ground, it moves in a horizontal direction at a constant rate, since no forces act in this direction after it is released.\ \DTN\ But the object will accelerate do  4;Pek~t ҷ}:$of ~h!$d3scribes an unusual type of acceleration in the section on Simple Harmonic Motion. Disk #2, Momentum and Work, introduces additional types of vectors in the section on Work.\ \BESC\ \L\ONE DONE\ \BESC\ \BL\FORCE\ \L\VOCAB\ \RVR\0\ \BESC\ \L\REVIEW\ \RRQ\0\ \BESC\ \L\MORE\ \DCW\ \SCH\2\2\ \DTCNW\More On This Topic\ \SCH\1\1\ \DTN\ All of the Physics disks deal with distances and time. Disk #4, Waves and Sound, deion constant to have it land in 500 m? How must you change the initial horizontal velocity to have it land in 500 m?\ \L\EXPBAL\ \DCW\ \SCT\0\5\ \DTNNW\ Do you want to run this simulation now? \ \GYN\ \BNO\ONE DONE\ \L\ONE\ \R\PDISK1\ ork, introduces additional types of vectors in the section on Work.\ \BESC\ sual type of acceleration in the section on Simple Harmonic Motion. Disk #2, Momentum and Work, introduces additional types of vectors in the section on Work.\ \BESC\ BESC\ \BL\FORCE\ \L\VOCAB\ \RVR\0\ \BESC\ \L\REVIEW\ \RRQ\0\ \BESC\ \L\MORE\ \DCW\ \SCH\2\2\ \DTCNW\More On This Topic\ \SCH\1\1\ \DTN\ All of the Physics disks deal with distances and time. Disk #4, Waves and Sound, describes an unu have it land in 500 m? How must you change the initial horizontal velocity to have it land in 500 m?\ \L\EXPBAL\ \DCW\ \SCT\0\5\ \DTNNW\ Do you want to run this simulation now? \ \GYN\ \BNO\ONE DONE\ \L\ONE\ \R\PDISK1\ \L\ONE DONE\ \ct.\ \DTN\ For a challenge, try to find conditions that cause the object to land 1000 m from the starting point. How must you change the initial height to have it land in 500 m? How must you change the gravitational acceleration constant to on the following three factors: - initial height - gravitational acceleration constant - initial horizontal velocity\ \DTN\ In this simulation, you may vary these features and see the effects of these changes on the path of the falling obje If the mass of an object is 10 kg, the gravitational force is 9.8 x 10 = 98 N.\ \DTN\ This factor, 9.8 N/kg, is the gravitational acceleration constant.\ \DTN\ How far the object will move horizontally before it hits the ground will dependwnward as it falls to the ground.\ \DTN\ The time it takes to fall depends on the height at which it is released and the magnitude of the gravitational force acting on it.\ \DTN\ On the surface of the earth, the gravitational force is 9.8 N/kg. 666?? >< 3 30 +.   000 - - ? ? 0 3;733  30??003??003<33?0 333333>0    0 3 HȱhlȱȱHȱ h` P X Ҧf* ӌڌC >@Ϥ Q RϠ © `   `A[i  A[i ʰ۠ Ȅ`     ߏL`l K 9786 H66Аl86`Ϣ   )   `Ϣ  )  ) a{iઘ)* &  e 8 ߏ `)  ߏ  i   )0" H $(*0ޠ L" A©@é~v ϮϢ  )) )c )LK8LL $ȥ ȥ Mȩȅ  ) ȱ  )Li i e WLQW{" `W ȑLQW ȑLQ Q Ɍ &Q ȱ LmWWiLX`  ȱp00LWL Q_ou HH ` ) )Ƀȑ `#ʊH $(*)+!hL8W% qqW{" X`XL ȱ L%"HH "LL%"HH "LHH LLH H L QENQbReTyWYX` ȱ L6%p0L8%%" ` ȑ LՋ%p0L8%%" `!$$p00L#ee%p00LH H L%$" W`W$" `X WLW%$" L Q  , LQXXLBW% qqW$" X`XL ȱ Lъ%" "LL%"#ee%" H H "%"" L LLB L QEXQgRjTWYŠX` ȱ 666666$B~0>II>66666 <66c<<66fnvf&66f8l <66l<,2< 2*&<8$<  0(< 0< $8< $$$8 >~,,,,,&&&&&>>"">$BB$$f2*&<8$<  0(< 0< $8< $$$8 s?33 ~~ |   0>3>333>>00>333>3?>6>33>03333  33 ?511333333333>33>003>063333>333 115?;3 3333>0? ?     $$ 15=> 33?3333333333333???>;3>333?333? ?00000333?1;?51113337;33333333333336333303? 33333333333 1115?;133 3333 ?0 ??? 0?00000? 3? `e e `ȥ 2 e e  e e  `  `   e  `   ` Ȅ`  `hh