8L2C)pJJJJ IH(ȱH:=IH[H`@HcH  $ +   I/H`JLNGȄBȄF aK  haaFF  mJm# KKJ UJ )J ۈ) ;J3ȱJFȱJGJKaȄM  aaNNJFLGJL L,8@E!x $ !DD<~~>~F=|q| |lL 3pxqss3s630300s666spp@gLLL GlllG`006cOglllg`  3OX@>ccc>?$$PDISK2.CVLz&$$PDISK2.CRV{$-SCICTRL.TXT| $$PDISK2.LST --PDISK2$' *PDISK2.FIG--PDISK2.LBL$-PDISK2.LLC$-*PDISK2.CCP`J$-PDISK2.CLCs$-*PDISK2.CVCw꽌ɪ꽌ɖ (I%ɠ@*<%<(=Ю@AKV<Y<<Y<&，YЬV0&^*^*&+`L ::RF? &PRODOS `DaElH$?EGvѶK+`L HHLy XP LM ŠϠĠӠS)*+,+`F)) (*=GJFjJJA QE'+ '== `@ STSP8QSS8 m P o R(8Rf<11|FFFvf|~F(;f y|AyMyx x x08000xs{qp@@AAAs3x133ax xx3630cACyLlllG`CcccG`sfffFG@@@@OX@XO  00?111?113c>>cc~`>  !~x`p~?@%%% ||,|| U *  U}}} U *j~?+ * Uu_UUU U jjjzoj j `| p~?  `|  ~x`p~?@@@ `@@`?oO   ![ px ||`0|| xppXXXp**""||,||U* U}} U * * U U j j    0pp`@@`pp0  %%% ||,|| U *  U}}} U *j~?+ * Uu_U jjjzoj `| p~?`|**""UUDD**UU**""UUDD**UU**""UUDD**UU**""UUDD**UU**""UUDD**UU**""UUDD**UU**""UUDD**UU**""UUDD**UUwNx `x?||nvn<\{8>`6  `x||nvn<\8`?{>`x||mum<\8`?{>> `x||nvn<\8`?{>@pxx\l\x8p@~w|C(PP@@**UU**""UUDD**UU@@`}`~`}@w@;g| p|~?~?w{w^?n=p @pxx\l\x~8wp|@ ~@``p~0p~`{`]@s~ x~@{?@}?@{?ofsllssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmmssmlssl ?`pp8X8p}pn`y? |nffL33Aw3LL3ff}_ff3LL?L33LfffL333LL3ffff3LLL33LfffL333LL3ff`f` o|nL33Lf ?Gf| nffL33`>8x8~n3LL3ffGG~ |88p}nff3LL | nL33Lfx`xFFn.,<nff3LL~Gc~x``c#'&6 nL33Lf@``pp``@nffL33 cAAc?n3LL3ff80~ nff3LLcAAc? @`p`@~~n3LL3ff@@?3!331`@`1nff3LL? <<6#?nL33Lfo|8nffL33?xp}`||n3LL3ff@`@ffL333LL3ffff3LLL33LfppffL33 x` x3LL3ff GGfn<<ff3LL |L33Lf@``pp``@ffL33@p`1q1117>GlllL>B(`ffyy66996699669966996699669966996699669966996699669966996699669966996699669966996699669966996699669966996699669966996699669966996699669966996699669966996699669966996699669yvff3LLL33Lf |0||FF||FFF|>cc>| |AxMxx``aaax 0 qs0x111cps(CACFG``p@p CcccG a111aGOXXXO@GlllG111XXXOF~>`??c1 9|L L|1ppxppf|` <@@@@3<f<<f<@~@p@p@p@p@~@p@p@p@p@|@`0` # %%%%@`@` % @@@% 00?111?%  <%||@~% }pxfff~``xpq<q3337<GlllL< X@X<1<000x |LLLx <`00`Mon Jan 21 12:30:30 1991 \DF\P2CREDS\1\1\1\0\ \DFB\PCREDS\25\100\ \XDEL\25\ \SVN\99\"0"\ \BL\NO MAINMENU\ \L\MAINMENU\ \SVN\99\"1"\ \L\NO MAINMENU\ \SSTW\1\0\39\0\23\ \L\MOMENTUM & WORK\ \SM\ \SMT\2: Momentum & Work\ \SMO\Work\ \SMO\Energy pcredsp2creds#ball11twoballs@twoballs2Kwork0Vwork1awork2lwork3wvary1vary2vary3animballpow1pow2momcons1momcons2momcons3momcons4momcons5smallballll\ \L\pcreds\ \L\p2creds\ \L\ball1\ \L\twoballs\ \L\twoballs2\ \L\work0\ \L\work1\ \L\work2\ \L\work3\ \L\vary1\ \L\vary2\ \L\vary3\ \L\animball\ \L\pow1\ \L\pow2\ \L\momcons1\ \L\momcons2\ \L\momcons3\ \L\momcons4\ \L\momcons5\ \L\smallba0pp`@@`pp00pp`@@`pp0 !xx    ~w|0p0@d    & ) ~~~~ `85  `88`8;`>   F  p|~~w{w^np???= . ...0@p|~~w{w^np@  @`0p@HPPQ$(0'????=PP($" @O p@"Kq}~~w{w^ns "x@?ax ???=3,p 0>3~(@p~F%@p```p@`0xx y"@p qxq@p#@p&@p)  @x,@@@`0@0`g[[CC@x/0`aaa1`82``@ @`@`@`@ l   g   f f  l @      l<|  A`8>`8;`8 pp8`< GG5`< <CcCg2p_/p?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrtu}~stance the object moved (0.6 m).\ \SQANS\4.8\ \SQLAST\No, (8 N) x (0.6 m) = 4.8 J.\ \SQA\T\CONTAINS\4.8\Right. (8 N) x (0.6 m) = 4.8 J.\ \SQA\F\=\48\No, you've got the idea, but you have to watchyour decimal point!\ \SQA\F\=\4800\No! The distanc240 J.\ \SQA\T\=\240\Right. (20 N) x (12 m) = 240 J.\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQF\S\5\ \SQM\3\ \SQT\When an 8 N force pushes an object 600 mm, ____ J of work are done.\ \SQH\Remember, W = Fd.\ \SQH\Multiply the force (8 N) by the di \SQ\1\Y\ \SQF\N\5\ \SQM\3\ \SQT\When an 20 N force pushes an object 12 m, ____ J of work are done.\ \SQH\Remember, W = Fd.\ \SQH\Multiply the force (20 N) by the distance the object moved (12 m).\ \SQANS\240\ \SQLAST\No, (20 N) x (12 m) = object 6 m, ____ J of work are done.\ \SQH\Remember, W = Fd.\ \SQH\Multiply the force (8 N) by the distance the object moved (6 m).\ \SQANS\48\ \SQLAST\No, (8 N) x (6 m) = 48 J.\ \SQA\T\=\48\Right. (8 N) x (6 m) = 48 J.\ \DQ\ \BQESC\ \DCW\e is the amount of work done moving an object 1 meter while applying a force of 1 newton.\ \DTN\ The unit is abbreviated J, so the work done moving the block was 147 J.\ \DCW\ \SQ\1\Y\ \SQF\N\5\ \SQM\3\ \SQT\When an 8 N force pushes an that:\ \SCH\2\3\ \DTCNW\W = Fd\ \SCH\1\1\ \DTNNW\we can calculate the work:\ \SCH\2\3\ \DTC\W = 29.4 N x 5 m = 147 J\ \SCH\1\1\ \DTN\ The work done moving this block was 147 joules.\ \DTN\ The unit of work is the joule. One joulelocity. During this period of constant velocity, the net force on the object must have been zero. If the frictional force was 29.4 N from right to left, the applied force must have been 29.4 N from left to right.\ \DCW\ \DTNNW\ Remembering\ \DFB\WORK1\126\10\ \DFB\WORK1\140\10\ \DFB\WORK1\154\10\ \DFB\WORK1\168\10\ \DFB\WORK1\182\10\ \DCW\ \DTN\ In this example, the block was pulled a distance of 5 meters.\ \DTN\ After an initial acceleration, the block moved at a constant vliding friction, , is 0.30.\ \DTN\ So the frictional force, F = F F = 0.30 x (10 kg x 9.8 m/s2) = 29.4 N\ \DTN\ Since the block moved from left to right, the frictional force was from right to left.\ \DFB\WORK1\98\10\ \DFB\WORK1\112\10ORK1\0\10\ \DFB\WORK1\14\10\ \DFB\WORK1\28\10\ \DFB\WORK1\42\10\ \DFB\WORK1\56\10\ \DFB\WORK1\70\10\ \DFB\WORK1\84\10\ \DFB\WORK1\98\10\ \SWT\\\6\\ \DTN\ Imagine a 10 kg block of wood being pulled along a board.\ \DTN\ The coefficient of s applied force, work is done.\ \DTN\ The amount of work depends on the force applied and the distance the object moves: \ \SCH\2\3\ \DTC\W = Fd\ \SCH\1\1\ \DTNNW\ \ \DTC\ Work = force x distance\ \DCW\ \DFB\WORK0\0\37\ \DFB\WC\2\WORK VECTOR\ \BMC\3\WORK VARY\ \BMC\4\WORK VOCAB\ \BMC\5\WORK REVIEW\ \BL\MOMENTUM & WORK\ \L\WORK INTRO\ \SCH\2\2\ \DTCNW\Introduction\ \SCH\1\1\ \DTN\ When a force applied to an object causes that object to move in the direction of the\VOCAB\ \BMC\7\REVIEW\ \BMC\8\MORE\ \BS\NETWORK\ \L\WORK\ \SM\ \SMT\Work\ \SMO\Introduction\ \SMO\Force Vectors\ \SMO\Varying Forces\ \SMO\Vocabulary Review\ \SMO\Review Questions\ \SMO\Return to Main Menu\ \DM\ \BMC\1\WORK INTRO\ \BMe is 600 mm, not 600 m. You must convert the millimeters to meters before multiplying.\ \DQ\ \BQESC\ \DCW\ \SQ\1\Y\ \SQF\N\2\ \SQM\3\ \SQT\When a 5 kg bar of chocolate is lifted 2 meters from the floor to a mouth, ____ J of work are done.\ \SQH\Remember that the force required to lift an object is gotten by multiplying the mass times the gravitational acceleration constant, 9.8 m/s2.\ \SQH\Multiply the force (5 x 9.8 N) by the distance the object is lifted (2 m).\ \SQANS\98\ \SQ \SQH\F = Fcos() = (15 N)  cos(40) = (15 N)  (.766)\ \SQANS\b\ \SQLAST\No, F = Fcos() = (15 N)  cos(40) = (15 N)  (.766) = 11.5 N\ \SQA\T\=\b\Right! Only about 3/4 as much force is needed.\ \DQ\ \SQSEND\ \BESC\ \L\WORK Vapplied was 15 N, how much force would have to be applied parallel to the surface to cause the same movement? a. 9.6 N c. 12.6 N e. 19.6 N b. 11.5 N d. 17.9 N f. 23.3 N\ \SQH\Remember, the parallel force is given by F = Fcos()\arallel to the direction of movement.\ \SQA\F\=\a\No, the force component is F cos(), not F sin()!\ \SQA\F\IS CONTAINED IN\ef\No, you want Fdcos(). Don't divide, multiply!\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQF\S\1\ \SQT\If the actual force c. 101 J f. 187 J\ \SQH\W = Fdcos()\ \SQANS\b\ \SQLAST\No, the work = Fdcos() = 15 x 8 x 0.766 = 92 J\ \SQA\T\=\b\Right. W = Fdcos() = 15 x 8 x 0.766 = 92 J.\ \SQA\F\=\d\No! You have to use the component of the force p66 tan(40) = .839\ \SCH\\1\ \SWT\\\R1\\ \SQ\ \SQF\S\1\ \SQT\A force of 15 N is applied to a block at an angle of 40. How much work is done by this force in sliding the block 8 m? a. 77 J d. 120 J b. 92 J e. 157 J the force is applied at a 30 angle from the plane,  W = 20 N  5 m  cos(30)   = 20 x 5 x 0.866   = 87 J \ \SWS\1\ \DCW\ \SQS\1\Y\ \SCH\\3\ \DTNNW\sin(40) = 0.643 cos(40) = 0.7e direction of movement.\ \DFB\WORK3\98\0\ \DCW\ \SCH\2\3\ \DTC\F = Fcos()\ \SCH\1\1\ \DTN\ where is the angle between the applied force and the direction of movement. \ \DCW\ \SCH\2\3\ \DTC\W = Fdcos()\ \SCH\1\1\ \DTN\ Since moving the block is, of course, given by the formula:\ \SCH\2\3\ \DTC\W = Fd\ \SCH\1\1\ \DTN\ But the force indicated in this formula is not the full 20 N applied to the block. It is only the component of the force that is parallel to thct.\ \DFB\WORK2\0\0\ \DFB\WORK2\14\0\ \DFB\WORK2\28\0\ \DFB\WORK2\42\0\ \DFB\WORK2\56\0\ \DFB\WORK2\70\0\ \DFB\WORK2\84\0\ \DFB\WORK2\98\0\ \DTN\ As a result of this applied force, the block slides 5 m to the right.\ \DTNNW\ The work done force vector that is parallel to the direction of movement will contribute to the work of moving the object.\ \DCW\ \DFB\WORK0\0\37\ \DFB\WORK2\0\0\ \SWT\\\7\\ \DTN\ A force of 20 N, directed up and to the right, is applied to this 10 kg objetimes the gravitational acceleration constant, 9.8 m/s2.\ \DQ\ \BESC\ \L\WORK VECTOR\ \SCH\2\2\ \DTCNW\Force Vectors\ \SCH\1\1\ \DTN\ If the force applied to an object is not parallel to the direction of movement, only the component of thex@`@`0'<`.p6x@9<`;p=x@@<`Bp Dx@ G<`IpI 1ox@xME@`0 **x@ @`0 UU<`@`0  **p @`0 UUx@p@`0 ** <` UU pv$Q&**)&UU)&**)&UU)&**)&UU)&**)&UU)&**)&UU)&**)&UU)&**)&UU)&**)"P@UU@P%"@@`bjjb@@&`x~ ~x`"s_;/{_wW?7Wy"@`0 Vv{{}}{xxpp`@@`0x~ >x@`0 |7.k~W]_uo-?u @`0LAST\No, (5 kg) x (9.8 m/s2) x (2 m) = 98 J\ \SQA\T\=\98\Right. (5 kg) x (9.8 m/s2) = 49 N, and (49 N) x (2 m) = 98 J.\ \SQA\F\=\10\No! The force required to lift the chocolate is not 5 N. The force is gotten by multiplying the mass (5 kg) ARY\ \SCH\2\2\ \DTCNW\Varying Forces\ \SCH\1\1\ \DTNNW\ To determine the amount of work done moving an object, you use the formula:\ \SCH\2\3\ \DTC\W = Fd\ \SCH\1\1\ \DTN\ This will not work, unfortunately, if the force varies during the course of the movement.\ \DF\VARY1\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ This graph shows how the force applied to an object varies as the object moves along a surface.\ \DTN\ In this example, the force doubles from 5 N to 10 N after the object movenergy cannot be used up.\ \DTN\ Instead, it is transformed into a different form.\ \DTN\ While there are many different forms of energy, two important types are kinetic energy and potential energy.\ \DCW\ \DTN\ Kinetic energy is enes something is energy.\ \DTN\ Energy is defined as "the ability to do work."\ \DTN\ The amount of energy in a system always remains constant.\ \DCW\ \DTN\ If the amount of energy in a system does not change, then when work is done, the ERGY CONSERVE\ \BMC\3\ENERGY POTENTIAL\ \BMC\4\ENERGY KINETIC\ \BMC\5\ENERGY VOCAB\ \BMC\6\ENERGY REVIEW\ \BL\MOMENTUM & WORK\ \L\ENERGY INTRO\ \SCH\2\2\ \DTCNW\Introduction\ \SCH\1\1\ \DTN\ When work is done, something must DO the work. ThiBESC\ \L\ENERGY\ \SM\ \SMT\Energy\ \SMO\Introduction\ \SMO\Conservation of Energy\ \SMO\Potential Energy\ \SMO\Kinetic Energy\ \SMO\Vocabulary Review\ \SMO\Review Questions\ \SMO\Return to Main Menu\ \DM\ \BMC\1\ENERGY INTRO\ \BMC\2\ENpezoid is A = b x (h1 + h2)/2\ \SQANS\6\ \SQLAST\No, W = (1/2) x (2 m) x (2 N + 4 N) = 6 J.\ \SQA\T\=\6\Right. W = (1/2) x (2 m) x (2 N + 4 N) = 6 J.\ \DQ\ \SQSEND\ \BESC\ \L\WORK VOCAB\ \RVR\1\ \BESC\ \L\WORK REVIEW\ \RRQ\1\ \\QH\ \SQT\How much work was done moving it the last two meters?\ \SQH\Calculate the area under the curve between the points d = 2 m and d = 4 m.\ \SQH\Calculate the area of the trapezoid with base 2 and heights 2 and 4.\ \SQH\The area of a tra and height 4.\ \SQH\The area of a triangle is A = 1/2 (base x height)\ \SQANS\8\ \SQLAST\No, the work = (1/2) x (4 m) x (4 N) = 8 J.\ \SQA\T\=\8\Right. The work = (1/2) x (4 m) x (4 N) = 8 J.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQF\N\5\= 0.5 J.\ \DQ\ \BQESC\ \DCW\ \SQ\ \SQF\N\5\\QH\ \SQT\Moving it the entire 4 meters, ____ J of work were done.\ \SQH\Calculate the area under the curve between the points d = 0 m and d = 4 m.\ \SQH\Calculate the area of the triangle with base 4 a triangle is A = 1/2 (base x height)\ \SQANS\0.5\ \SQLAST\No, the work = (1/2) x (1 m) x (1 N) = 0.5 J.\ \SQA\F\IS CONTAINED IN\0.\No, it has to take SOME work!!\ \SQA\T\IS CONTAINED IN\0.5\Right. The work = (1/2) x (1 m) x (1 N) \SQF\S\3\\QH\ \SQT\To move this block the first meter required ____ J of work.\ \SQH\Calculate the area under the curve between the points d = 0 m and d = 1 m.\ \SQH\Calculate the area of the triangle with base 1 and height 1.\ \SQH\The area of\R2\\1\ \DTN\ The force required to stretch a spring increases as a function of the length of the spring. This graph shows the force required to move a 5 kg block attached to a spring.\ \SQS\1\Y\ \DF\VARY3\1\0\1\0\ \SWT\\\R2\\1\ \DCW\ \SQ\ work done is equal to the area under the curve. In this particular case this is clearly true because the area of each rectangle is the height x base = force x distance, which is the definition of the work carried out.\ \DF\VARY3\1\1\1\0\ \SWT\\ = Fd = (10 N) x (6 m) = 60 J\ \DTN\ Total work (0 - 10 m): W = W1 + W2 = (20 J) + (60 J) = 80 J\ \DF\VARY2\1\0\1\0\ \SWT\\\R2\\1\ \DCW\ \DTN\ Notice that for each section of this curve, thees 4 m.\ \DTN\ To calculate the total work done, you must calculate the amounts of work done at each force and then add them.\ \DCW\ \DTN\ 0 - 4 m: W1 = Fd = (5 N) x (4 m) = 20 J\ \DTN\ 4 - 10 m: W2rgy in the form of motion.\ \DTN\ When a car is moving down the road, it has a large amount of kinetic energy. If it hits a stationary object, it will exert a force on that object and cause it to move, thereby doing work.\ \DTN\ When a hammer hits a nail (or a thumb), the moving hammer exerts a force on the object it hits and causes it to move, doing work.\ \DTN\ In both of these examples, the moving object stops as the result of a collision. When it stops, the kinetic energy of gravitational potential energy.\ \DTN\ The gravitational potential energy of an object is defined as the amount of kinetic energy the object would gain in falling to the ground.\ \DTN\ Since the amount of potential energy gained is equal and matter cannot enter a closed system.\ \DQ\ \BESC\ \L\ENERGY POTENTIAL\ \SCH\2\2\ \DTCNW\Potential Energy\ \SCH\1\1\ \DTN\ When work is done to lift an object, energy is expended. This energy is stored in the object itself, in the form er cannot enter a closed system.\ \SQH\Remember the definition of a closed system.\ \SQA\T\CONTAINS\y\No it isn't! Water is entering the bathtub, and matter cannot enter a closed system.\ \SQA\T\CONTAINS\n\Right. Water is entering the bathtub, stem for energy, since sunlight can enter the system, heating the interior and adding energy to the box.\ \DCW\ \SQ\2\Y\ \SQF\S\3\ \SQT\Is a bath tub filling with water a closed system?\ \SQANS\no\ \SQLAST\No, water is entering the tub and matt The phrase "a closed system" in these definitions means an area where no energy or matter enters or leaves.\ \DTN\ A sealed glass box is a closed system with regard to matter, since nothing can get in or out.\ \DTN\ But it is not a closed syy.\ \DTN\ Thus, the concepts of conservation of energy and conservation of matter have been supplanted by the "Law of Conservation of Energy and Matter," which states that the sum of the matter and energy in a closed system is constant.\ \DTN\ aying constant and its energy as increasing, or you may simply think of its mass as increasing.\ \DTN\ The relativistic mass of the moving ball is greater than its resting mass because it includes the mass-equivalent of the ball's kinetic energial energy and kinetic energy are different forms of energy, so must energy and matter be thought of as different forms of the same quantity.\ \DTN\ According to the Theory of Relativity, when you throw a ball, you may think of the ball's mass as stse. They also believed that the amount of matter in a closed system could neither increase nor decrease.\ \DTN\ But when Albert Einstein presented his Theory of Relativity in 1905, this view was rejected. Einstein said that just as potentts a kettle.\ \DG\ \BESC\ \L\ENERGY CONSERVE\ \SCH\2\2\ \DTCNW\Energy Conservation\ \SCH\1\1\ \DTN\ Until the start of the twentieth century, physicists believed that the amount of energy in a closed system could neither increase nor decreakinetic energy into a different form?\ \SGO\|Y An arrow strikes a target.\ \SGO\|Y A hammer hits a nail.\ \SGO\|Y A pickax breaks a stone.\ \SGO\A battery turns a motor.\ \SGO\A battery lights a light.\ \SGO\Sunlight warms a puddle.\ \SGO\A fire hea the book falls, it accelerates, and the potential energy is converted back into kinetic energy. If the book lands on your toe, it will again do work, much to your dismay.\ \SQG\2\Y\ \SGT\Which of these situations reflect the transformation of rom the floor up onto a table, the energy is transformed into potential energy.\ \DTN\ The book now contains potential energy. This potential energy can be converted to kinetic energy by sliding the book off the edge of the table.\ \DTN\ Asof the object is transformed into other forms of energy.\ \DCW\ \DTN\ Potential energy is stored energy, which is available for doing work, but which is not in the form of motion.\ \DTN\ When work is done (and energy used) to lift a book fto the amount of work done lifting the object, we know that:\ \SCH\2\3\ \DCW\ \DTCNW\E = W = Fd\ \SCH\1\1\ \DTN\where E is the potential energy.\ \DTN\ In the case of an object lifted into the air, the distance, d, is just the height the object was lifted, h. The force, F, is equal to the mass of the object, m, times the gravitational acceleration constant, g.\ \DTNNW\So:\ \SCH\2\3\ \DTC\E = mgh\ \SCH\1\1\ \DTN\ When the mass is in kg, the acceleration in m/s2, and th\ \SWT\\\4\\ \DTN\ If the plane is not frictionless, more work must be done to push the block up the incline.\ \DTN\ This energy is not, however, transformed into potential energy. Instead, it is converted into heat.\ \DTN\ This is why g,\ \SCH\2\3\ \DTC\W = mgdsin()\ \SCH\1\1\ \DTNNW\ But dsin() = h, the height the objects rises, so:\ \SCH\2\3\ \DTC\W = mgh\ \SCH\1\1\ \DTN\ Thus, the potential energy gained by the object is indeed equal to the work done.\ \DCWh = d sin() = 12 x 0.500\ \DQ\ \SQSEND\ \DCW\ \DTNNW\ When an object is pushed up a frictionless inclined plane, the work done on it is:\ \SCH\2\3\ \DTC\W = Fd = [Fsin()]d\ \SCH\1\1\ \DTNNW\ Since F is the gravitational force, m x = 392 J\ \SQA\T\=\294\Right, W = (5 kg)x(9.8 m/s2)x(6 m)\ \SQA\F\=\290\More accuracy, please.\ \SQA\F\=\300\More accuracy, please.\ \SQA\F\=\588\No, h = d sin() = 12 x 0.500\ \SQA\F\=\590\No, h = d sin() = 12 x 0.500\ \SQA\F\=\600\No, sed 30 above the horizontal, it gains ____ J of potential energy.\ \SQH\Remember, W = Fd = mgh\ \SQH\Remember that the height the object is lifted is only 12 sin() = 12 x 0.500 = 6 m.\ \SQANS\294\ \SQLAST\No, W = (5 kg)x(9.8 m/s2)x(6 m) 5 kg and the height is 0.2 m.\ \DQ\ \BQESC\ \DCW\ \SQS\2\Y\ \SCH\\3\ \DTNNW\ sin(30) = 0.500 cos(30) = 0.866 tan(30) = 0.577\ \SWT\\\R1\\ \SCH\\1\ \SQ\ \SQF\N\5\ \SQT\When a 5 kg object is pushed 12 m up a frictionless inclined plane raiSQANS\0.98\ \SQLAST\No, W = (2 kg)x(9.8 m/s2)x(5 m) = 0.98 J.\ \SQA\T\IS CONTAINED IN\0.98\Right, W = (2 kg)x(9.8 m/s2)x(5 m) = 0.98 J.\ \SQA\F\IS CONTAINED IN\1.00\More accuracy, please.\ \SQA\F\IS CONTAINED IN\98.00\No, the mass is 0. m/s2.\ \DQ\ \BQESC\ \DCW\ \SQ\2\Y\ \SQF\S\5\ \SQT\When a 500 g object is lifted 200 mm, it gains ____ J of potential energy.\ \SQH\Remember, W = Fd = mgh\ \SQH\Remember that the mass, 500 g, is 0.5 kg and the height, 200 mm, is 0.2 m.\ \s2.\ \SQANS\392\ \SQLAST\No, W = (5 kg)x(9.8 m/s2)x(8 m) = 392 J\ \SQA\T\=\392\Right, W = (5 kg)x(9.8 m/s2)x(8 m)\ \SQA\F\=\390\More accuracy, please.\ \SQA\F\=\400\More accuracy, please.\ \SQA\F\=\40\No, W = Fd, but F = 5 kg x 9.8.8 m/s2.\ \DQ\ \BQESC\ \DCW\ \SQ\2\Y\ \SQF\N\5\ \SQT\When a 5 kg object is lifted 8 m, it gains ____ J of potential energy.\ \SQH\Remember, W = Fd = mgh\ \SQH\Remember that the gravitational acceleration constant, g, is equal to 9.8 m/acceleration constant, g, is equal to 9.8 m/s2.\ \SQANS\98\ \SQLAST\No, W = (2 kg)x(9.8 m/s2)x(5 m) = 98 J\ \SQA\T\=\98\Right, W = (2 kg)x(9.8 m/s2)x(5 m)\ \SQA\T\=\100\More accuracy, please.\ \SQA\F\=\10\No, W = Fd, but F = 2 kg x 9 \SQF\N\5\ \SQT\When a 2 kg object is lifted 5 m, it gains ____ J of potential energy.\ \SQH\Remember, the potential energy gained is equal to the work done lifting the object.\ \SQH\Remember, E = mgh\ \SQH\Remember that the gravitational s2.\ \SQANS\98\ \SQLAST\No, W = (2 kg)x(9.8 m/s2)x(5 m) = 98 J\ \SQA\T\=\98\Right, W = (2 kg)x(9.8 m/s2)x(5 m)\ \SQA\F\=\100\More accuracy, please.\ \SQA\F\=\10\No, W = Fd, but F = 2 kg x 9.8 m/s2.\ \DQ\ \BQESC\ \DCW\ \SQ\2\Y\e height in m, the energy is in joules.\ \DCW\ \SQ\2\Y\ \SQF\N\5\ \SQT\When a 2 kg object is lifted 5 m, ____ J of work is done.\ \SQH\Remember, W = Fd = mgh\ \SQH\Remember that the gravitational acceleration constant, g, is equal to 9.8 m/friction always heats up the objects that are rubbing together.\ \BESC\ \L\ENERGY KINETIC\ \SCH\2\2\ \DTCNW\Kinetic Energy\ \SCH\1\1\ \DTN\ Moving objects have kinetic energy and hence are capable of doing work.\ \DTNNW\ The amount of energy depends on the mass and velocity of the object:\ \SCH\2\3\ \DTC\ E = 1/2mv2\ \SCH\1\1\ \DTN\ If the mass is in kg and velocity is in m/s2, the energy is in joules.\ \DTN\ Imagine a 5 kg object that falls from a height of 19.6 m.\ ther common unit of power is BTU/hr. The BTU, or British Thermal Unit, is defined as the amount of heat required to raise the temperature of 1 lb of water 1 degree Fahrenheit.\ \DTN\ A furnace with a power rating of 30,000 BTU/hr produces heat es of electrical energy into light and heat every second.\ \DTN\ Another common unit of power is the horsepower. A one horsepower engine has a power of 746 watts. It is capable of carrying out 746 joules of work per second.\ \DTN\ Anouivalent to one joule per second.\ \DTN\ The power of a machine indicates how rapidly it can do work. If a machine can do 100 J of work every second, it has a power of one hundred watts, or 100 W.\ \DTN\ A 100 W light bulb converts 100 joul & WORK\ \L\POWER INTRO\ \SCH\2\2\ \DTCNW\Introduction\ \SCH\1\1\ \DTN\ The term power refers to the rate at which work is done.\ \DTN\ If work is measured in joules and time in seconds, then power is measured in watts. 1 watt is eqPOWER\ \SM\ \SMT\Power\ \SMO\Introduction\ \SMO\Calculating Power\ \SMO\Vocabulary Review\ \SMO\Review Questions\ \SMO\Return to Main Menu\ \DM\ \BMC\1\POWER INTRO\ \BMC\2\POWER CALC\ \BMC\3\POWER VOCAB\ \BMC\4\POWER REVIEW\ \BL\MOMENTUM/ 2) = 1350 / 450 = 3 kg\ \SQA\T\=\3\Right. m = E / (1/2 v2) = 1350 / (30 x 30 / 2) = 1350 / 450 = 3 kg\ \DQ\ \BESC\ \L\ENERGY VOCAB\ \RVR\2\ \BESC\ \L\ENERGY REVIEW\ \RRQ\2\ \BESC\ \L\ \SQF\N\5\ \SQT\A bowling ball which has a velocity of 30 m/s and a kinetic energy of 1,350 J has a mass of ____ kg.\ \SQH\E = 1/2 mv2\ \SQH\m = E / (1/2 v2)\ \SQANS\3\ \SQLAST\No, m = E / (1/2 v2) = 1350 / (30 x 30 E = 1/2 mv2 = 2 x 5 x 5 / 2 = 25 J\ \SQA\T\=\25\Right. E = 1/2 mv2 = 2 x 5 x 5 / 2 = 25 J\ \SQA\F\=\50\No, E = 1/2|c mv2\ \SQA\F\=\5\No, E = 1/2 mv2\ \DQ\ \BQESC\ \DCW\ \SQ\2\Y\ SQA\F\=\72\No, E = 1/2 mv2\ \SQA\F\=\12\No, E = 1/2 mv2\ \DQ\ \BQESC\ \DCW\ \SQ\2\Y\ \SQF\N\5\ \SQT\A 2 kg mass moving at a speed of 5 m/s contains ____ J of kinetic energy.\ \SQH\E = 1/2 mv2\ \SQANS\25\ \SQLAST\No, vation of Energy.\ \DCW\ \SQ\2\Y\ \SQF\N\5\ \SQT\An 8 kg mass moving at a speed of 3 m/s has ____ J of kinetic energy.\ \SQH\E = 1/2 mv2\ \SQANS\36\ \SQA\T\=\36\Right. E = 1/2 mv2 = 8 x 3 x 3 / 2 = 36 J\ \entical!\ \DTN\ Thus the work done lifting an object, the potential energy it gains by being lifted, and the kinetic energy it gains by falling back to its starting point are all the same.\ \DTN\ This is a reflection of the Law of Conser2)(5)(19.62) = 960 J\ \DTN\ By comparison, the potential energy held by the object before it began to fall was E = mgh = (5)(9.8)(19.6) = 960 J\ \DTN\ The kinetic and potential energies are idvelocity of a falling object is given by v = at, so when it reaches the ground this object will have a velocity of (9.8)(2) = 19.6 m/s.\ \DTN\ Its kinetic energy on reaching the ground is given by E = (1/2)mv2 = (1/ \DTN\ Since the object was not moving when it was first released, the distance it falls will equal (1/2)at2 where a = 9.8 m/s2. Thus, t must equal 2 s.\ \DCW\ \DTN\ d = (1/2)(9.8)(22) = 19.6 m\ \SCH\1\1\ \DTN\ Remember that the at a rate equivalent to about 10,000 watts.\ \SQP\3\Y\ \SPT\Match each unit with its rate of work.\ \SPO1\watt\horsepower\BTU/hr\ \SPO2\ 1 J/s = 1 __________\ \SPO2\746 J/s = 1 __________\ \SPO2\0.3 J/s = 1 __________\ \DP\ \BESC\ \L\POWER CALC\ \SCH\2\2\ \DTCNW\Calculating Power\ \SCH\1\1\ \DTN\ How does one calculate the power exerted when work is being done?\ \DCW\ \DFB\WORK0\0\38\ \DFB\WORK1\0\10\8\ \DFB\WORK1\14\10\8\ \DFB\WORK1\28\10\8\ \DFB\WORK1\42\10\8\ \D\No, the work is the force (5 kg x 9.8 m/s2) times the distance (2 m), or 98 J.\ \SQA\T\=\98\Right. The work is the force (5 kg x 9.8 m/s2) times the distance (2 m), or 98 J.\ \SQA\F\=\100\More accuracy, please.\ \SQA\F\=\9\No, the maity is 3 m/s.\ \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQF\N\5\ \SQH\P = Fv\ \SQH\F = mg\ \SQH\P = Fv\ \SQH\F = mg\ \SQT\To lift a 5 kg loaf of bread 2 m into the air, ____ J of work must be done.\ \SQH\W = Fd\ \SQH\F = mg\ \SQANS\98\ \SQLAST Fv\ \SQANS\15\ \SQLAST\No, the power is the force (5 N) times the velocity (3 m/s).\ \SQA\T\=\15\Right! The power is the force (5 N) times the velocity (3 m/s).\ \SQA\F\=\6\No, P = Fv. The force is 5 N.\ \SQA\F\=\30\No, P = Fv. The veloc 5 N.\ \SQA\F\=\15\No, W = Fd. The distance is 6 m.\ \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQF\N\5\ \SQT\To push a 2 kg mass up an inclined plane a distance of 6 m at a speed of 3 m/s, with a force of 5 N. ____ W of power must be applied.\ \SQH\P =\ \SQANS\30\ \SQLAST\No, the work is the force (5 N) times the total distance the object is moved (6 m).\ \SQA\T\=\30\Right! The work is the force (5 N) times the total distance the object is moved (6 m).\ \SQA\F\=\12\No, W = Fd. The force isr is equal simply to the force times the velocity!\ \SWS\1\ \DCW\ \SQ\3\Y\ \SQF\N\5\ \SQT\A 2 kg mass is pushed up an inclined plane a distance of 6 m at a speed of 3 m/s, with a force of 5 N. A total of ____ J of work are done.\ \SQH\W = Fdw substitute this into the equation for power, we get:\ \DCW\ \SCT\1\0\ \SCH\2\1\ \DTNW\ W Fvt \ \DTNW\P = = = Fv\ \DTNW\ t t\ \XPLOT\49\85\ \XDRAW\40\0\0\1\-40\0\ \XPLOT\119\85\ \XDRAW\62\0\0\1\-62\0\ \DTN\ The poweow the velocity (v = 5 m/s) and the time (t = 5 s), we can now calculate the power.\ \DCW\ \DTNNW\ If we substitute the equation for distance (d = vt) into the one for work (W = Fd), we get:\ \SCH\2\3\ \DTC\W = Fvt\ \SCH\1\1\ \DTN\ If we noe know the time (t = 5 s) but we need to calculate the work.\ \SCH\2\3\ \DCW\ \DTCNW\W = Fd\ \SCH\1\1\ \DTN\ We know the force (F = 10 N) but we still need to calculate the distance.\ \SCH\2\3\ \DTCNW\d = vt\ \SCH\1\1\ \DTN\ Since we knW\ Since the power used to move the block is:\ \SCH\2\3\ \DTCNW\P = W/t,\ \SCH\1\1\ \DTN\the power used is just (100 J)/(5 s), or 20 W.\ \DCW\ \DTN\ To get this answer took three steps:\ \SCH\2\3\ \DTCNW\P = W/t\ \SCH\1\1\ \DTN\ W0\10\8\ \DFB\WORK1\14\10\8\ \DFB\WORK1\28\10\8\ \DFB\WORK1\42\10\8\ \DFB\WORK1\56\10\8\ \DFB\WORK1\70\10\8\ \DFB\WORK1\84\10\8\ \DFB\WORK1\98\10\ \SWT\\\7\\ \DCW\ \DTN\ In this example, 100 joules of work moved this block 10 m in 5 s.\ \DTNNtal distance the object is moved (2 m/s x 5 s).\ \SQA\F\=\20\No, W = Fd. The distance isn't 2 m, the velocity is 2 m/s.\ \SQA\F\=\50\No, W = Fd. The distance isn't 5 m, the time is 5 s.\ \DQ\ \SWS\1\ \DCW\ \DFB\WORK0\0\38\ \DFB\WORK1\ing the block.\ \SQH\Remember, W = Fd.\ \SQH\Remember, d = vt.\ \SQANS\100\ \SQLAST\No, the work is the force (10 N) times the total distance the object is moved (2 m/s x 5 s).\ \SQA\T\=\100\Right! The work is the force (10 N) times the toFB\WORK1\56\10\8\ \DFB\WORK1\70\10\8\ \DFB\WORK1\84\10\8\ \DFB\WORK1\98\10\ \SWT\\\7\\ \SQ\3\Y\ \SQF\N\5\ \SQT\A force of 10 N is used to push a block along a surface. The block moves at a velocity of 2 m/s for 5 s. ____ J of work are done movss is 1 kg, but the force is (1 kg) x (9.8 m/s2).\ \SQA\F\=\10\No, the mass is 1 kg, but the force is (1 kg) x (9.8 m/s2).\ \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQF\N\5\ \SQT\To lift a 5 kg loaf of bread 2 m into the air at the rate of 1 m/s, power must be applied at a rate of ____ W.\ \SQH\P = Fv\ \SQH\F = mg\ \SQANS\49\ \SQLAST\No, the power is the force (49 N) times the velocity (1 m/s), or 49 W.\ \SQA\T\=\49\Right. The power is the force (49 N) times the velocity (1 m/N\ The dot between kg and m in these units indicates that you get the momentum by multiplying the mass (in kg) times the distance (in meters). The slash indicates that you must then divide this product by the time (in seconds).\ \SQS\4\Yvelocity. Its direction is equal to the direction of its velocity.\ \DTN\ Momentum, like velocity, has no special name for its units. Velocity has the units meters per second, or m/s.\ \DCW\ \SCH\2\3\ \DTCNW\kgm/s\ \SCH\1\1\ \DTd velocity.\ \SCH\2\3\ \DTCNW\p = mv\ \SCH\1\1\ \DTN\where p stands for the momentum of the object.\ \DTN\ Like velocity, momentum is a vector quantity. Its magnitude is equal to the mass of the object times the magnitude of its ic energy can be changed into potential energy.\ \DTN\ But there is an aspect of movement which is conserved. This is the momentum of a moving object.\ \DCW\ \DTNNW\ The momentum of an object is defined as the product of its mass an the Conservation of Energy and Matter states that the total amount of matter and energy in the universe remains constant at all times.\ \DTN\ While the energy of a system remains constant, the kinetic energy is not necessarily conserved. Kinet \SMO\Return to Main Menu\ \DM\ \BMC\1\MOMENT INTRO\ \BMC\2\MOMENT IMPULSE\ \BMC\3\MOMENT CONSERVE\ \BMC\4\MOMENT VOCAB\ \BMC\5\MOMENT REVIEW\ \BL\MOMENTUM & WORK\ \L\MOMENT INTRO\ \SCH\2\2\ \DTCNW\Introduction\ \SCH\1\1\ \DTN\ The Law of = 6.04 m/s\ \DQ\ \BESC\ \L\POWER VOCAB\ \RVR\3\ \BESC\ \L\POWER REVIEW\ \RRQ\3\ \BESC\ \L\MOMENTUM\ \SM\ \SMT\Momentum\ \SMO\Introduction\ \SMO\Impulse\ \SMO\Conservation of Momentum\ \SMO\Vocabulary Review\ \SMO\Review Questions\s\ \SQH\F = mg = 617 N\ \SQH\P = 5 x 746 W = 3730 W\ \SQH\v = P/F\ \SQANS\c\ \SQLAST\No, the velocity = P/F = (5 x 746 W)/(70 kg x 9.8 m/s2) = 6.04 m/s\ \SQA\T\=\c\Right. The velocity = P/F = (5 x 746 W)/(70 kg x 9.8 m/s2) 8 m/s2).\ \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQF\S\1\ \SQT\One horsepower is equal to 746 W. At what velocity can a 5 hp engine lift a 63 kg adult? a. 0.08 m/s d. 11.8 m/s b. 1.21 m/s e. 30.2 m/s c. 6.04 m/s f. 59.2 m/686 W.\ \SQA\T\=\686\Right. The power is the force (686 N) times the velocity (1 m/s), or 686 W.\ \SQA\F\=\690\More accuracy, please.\ \SQA\F\=\700\More accuracy, please.\ \SQA\F\=\70\No, the mass is 70 kg, but the force is (70 kg) x (9.\ \SQF\N\5\ \SQT\To lift a 70 kg adult 100 m into the air at the rate of 1 m/s, power must be applied at the rate of ____ W.\ \SQH\P = Fv\ \SQH\F = mg\ \SQANS\686\ \SQLAST\No, the power is the force (686 N) times the velocity (1 m/s), or r is the force (49 N) times the velocity (1 m/s), or 49 W.\ \SQA\F\=\4\No, the mass is 5 kg, but the force is (5 kg) x (9.8 m/s2).\ \SQA\F\=\5\No, the mass is 5 kg, but the force is (5 kg) x (9.8 m/s2).\ \DQ\ \BQESC\ \DCW\ \SQ\3\Yof bread 12 m into the air at the rate of 1 m/s, power must be applied at a rate of ____ W.\ \SQH\P = Fv\ \SQH\F = mg\ \SQANS\49\ \SQLAST\No, the power is the force (49 N) times the velocity (1 m/s), or 49 W.\ \SQA\T\=\49\Right. The powes), or 49 W.\ \SQA\F\=\50\More accuracy, please.\ \SQA\F\=\5\No, the mass is 5 kg, but the force is (5 kg) x (9.8 m/s2).\ \SQA\F\=\98\No, P = Fv. The velocity is 1 m/s.\ \DQ\ \BQESC\ \DCW\ \SQ\3\Y\ \SQF\N\5\ \SQT\To lift a 5 kg loaf \ \DCW\ \SWT\\\4\\ \SA\ \DFA\BALL1\8\8\ \SAS\50\1\1\-1\ \SAL\26\10\0\0\0\ \DA\ \DTN\ Imagine that this animation showed a 2 kg ball moving a distance of 6 m, in 2 s.\ \SWT\\\R2\\1\ \SQ\4\Y\ \SQF\N\5\ \SQT\The velocity of the ball was ____ m/s.\ \SQH\d = vt\ \SQH\v = d/t\ \SQANS\3\ \SQLAST\No, v = d/t = (6 m)/(2 s) = 3 m/s.\ \SQA\T\=\3\Right. v = d/t = (6 m)/(2 s) = 3 m/s.\ \SQA\F\=\6\No, the distance is 6 m, but what is the velocity?\ \DQ\ \BQESC\ \DCW\ \SQ\4\Y\ \S use the mass of the wagon.\ \DQ\ \BESC\ \L\MOMENT CONSERVE\ \SCH\2\2\ \DTCNW\Conservation of Momentum\ \SCH\1\1\ \DTN\ The momentum of a system can neither increase nor decrease. This fact is stated by the law of Conservation of Momentums by a 20 N force is ____ kmm/s.\ \SQH\p = Ft\ \SQANS\200\ \SQLAST\No, p = Ft = (20 N) x (10 s) = 200 kgm/s\ \SQA\T\=\200\Right. p = Ft = (20 N) x (10 s) = 200 kgm/s\ \SQA\F\=\50\No, don't use the mass of the wagon.\ \SQA\F\=\100\No, don't = 147 kgm/s\ \SQA\T\=\150\More accuracy, please.\ \SQA\F\=\15\No, the force is (5 kg) x (9.8 m/s2)\ \DQ\ \BQESC\ \DCW\ \SQ\4\Y\ \SQF\S\11\ \SQT\The momentum of a 5 kg wagon after being pushed along a frictionless horizontal surface for 10 = Ft\ \SQH\F = mg\ \SQH\F = (5 kg) x (9.8 m/s2)\ \SQANS\147\ \SQLAST\No, p = Ft and F = (5 kg) x (9.8 m/s2) = 49 N, so p = (49 N) x (3 s) = 147 kgm/s\ \SQA\T\=\147\Right. p = Ft and F = (5 kg) x (9.8 m/s2) = 49 N, so p = (49 N) x (3 s)CH\1\1\ \DTN\ The momentum equals the force times the time.\ \DCW\ \SCH\2\3\ \DTCNW\p = Ft\ \SCH\1\1\ \SWT\\\R1\\ \SQ\4\Y\ \SQF\N\5\ \SQT\The momentum of a falling 5 kg rock, 3 seconds after it begins to fall, is ____ kmm/s.\ \SQH\pat, where a is the rate of acceleration.\ \DTNNW\ So its momentum is:\ \SCH\2\3\ \DTC\p = mv = mat\ \SCH\1\1\ \DTNNW\ But we already know that:\ \SCH\2\3\ \DTNNW\ F = ma\ \SCH\1\1\ \DTNNW\ so:\ \SCH\2\3\ \DTCNW\p = Ft\ \Sating at 2 m/s2, then its velocity after 1 s was 2 m/s and its momentum was:  p = mv = (5 kg) x (2 m/s) = 10 kgm/s \ \SWS\1\ \DCW\ \DTN\ More generally, after t seconds, the velocity of an object that is uniformly accelerating from rest is \SAL\28\0\1\0\0\ \DA\ \XERASE\234\8\247\18\ \SWT\\\6\\ \DTN\ The momentum gained by this ball during the first second depends on the mass of the ball and its velocity after one second.\ \DTN\ If the mass of the ball was 5 kg and it was acceler\DTCNW\Impulse\ \SCH\1\1\ \DTN\ When a force acts on an object in the absence of friction, the object accelerates. As it does so, its position, velocity, kinetic energy, and momentum all change.\ \DCW\ \SA\ \DFA\ball1\0\8\ \SAS\70\\\-1\ ing at a speed of 7 m/s is ____ kgm/s.\ \SQH\p = mv\ \SQANS\420\ \SQLAST\No, the momentum is mv = (60 kg)(7 m/s) = 420 kgm/s\ \SQA\T\=\420\Right. The momentum is mv = (60 kg)(7 m/s) = 420 kgm/s\ \DQ\ \BESC\ \L\MOMENT IMPULSE\ \SCH\2\2\ \SQH\p = mv\ \SQANS\350\ \SQLAST\No, the momentum is mv = (10 kg)(35 m/s) = 350 kgm/s\ \SQA\T\=\350\Right. The momentum is mv = (10 kg)(35 m/s) = 350 kgm/s\ \DQ\ \BQESC\ \DCW\ \SQ\4\Y\ \SQF\N\5\ \SQT\The momentum of a 60 kg woman runnmv = (2 kg)(3 m/s) = 6 kgm/s\ \SQA\T\=\6\Right. The momentum is mv = (2 kg)(3 m/s) = 6 kgm/s\ \DQ\ \SQSEND\ \BQESC\ \SWS\1\ \DCW\ \SQ\4\Y\ \SQF\N\5\ \SQT\The momentum of a 10 kg object falling at the speed of 35 m/s is ____ kgm/s.\ )2 = 9 J\ \SQA\F\=\18\No, E = 1/2 mv2\ \SQA\F\=\3\No, E = 1/2 mv2\ \DQ\ \BQESC\ \DCW\ \SQ\4\Y\ \SQF\N\5\ \SQT\The momentum of the ball was ____ kgm/s.\ \SQH\p = mv\ \SQH\v = d/t\ \SQANS\6\ \SQLAST\No, the momentum is QF\N\5\ \SQT\The kinetic energy of the ball was ____ J.\ \SQH\E = 1/2 mv2\ \SQH\v = d/t\ \SQANS\9\ \SQLAST\No, the energy is 1/2 mv2 = (1/2)(2 kg)(3 m/s)2 = 9 J\ \SQA\T\=\9\Right. The energy is 1/2 mv2 = (1/2)(2 kg)(3 m/s.\ \DTN\ When two balls collide, the sum of the momentums of the two balls is the same before and after the collision.\ \DTN\ Since momentum is a vector, it is the sums of the two momentum vectors before and after the collision that must remain constant.\ \DCW\ \DFB\TWOBALLS\0\27\ \DFB\TWOBALLS\7\27\ \DFB\TWOBALLS\14\27\ \DFB\TWOBALLS\21\27\ \DFB\TWOBALLS\28\27\ \DFB\TWOBALLS\35\27\ \DFB\TWOBALLS\42\27\ \DFB\TWOBALLS\49\27\ \DFB\TWOBALLS\56\27\ \DFB\TWOBALLS\63\27\ \DFB\POW2\63\0\ BALL\49\27\ \DFB\ANIMBALL\196\27\ \DFB\ANIMBALL\42\27\ \DFB\ANIMBALL\203\27\ \DFB\ANIMBALL\35\27\ \DFB\ANIMBALL\210\27\ \DFB\ANIMBALL\28\27\ \DFB\ANIMBALL\217\27\ \DFB\ANIMBALL\21\27\ \DFB\ANIMBALL\224\27\ \DFB\ANIMBALL\14\27\ \DFB\ANIMBALL\231\DFB\ANIMBALL\154\27\ \DFB\ANIMBALL\84\27\ \DFB\ANIMBALL\161\27\ \DFB\ANIMBALL\77\27\ \DFB\ANIMBALL\168\27\ \DFB\ANIMBALL\70\27\ \DFB\ANIMBALL\175\27\ \DFB\ANIMBALL\63\27\ \DFB\ANIMBALL\182\27\ \DFB\ANIMBALL\56\27\ \DFB\ANIMBALL\189\27\ \DFB\ANIMLL\182\27\ \DFB\ANIMBALL\70\27\ \DFB\ANIMBALL\168\27\ \DFB\ANIMBALL\84\27\ \DFB\ANIMBALL\154\27\ \DFB\ANIMBALL\98\27\ \DFB\ANIMBALL\140\27\ \DFB\POW2\105\0\ \DFB\ANIMBALL\140\27\ \DFB\ANIMBALL\98\27\ \DFB\ANIMBALL\147\27\ \DFB\ANIMBALL\91\27\ \y is transformed to heat energy.\ \DCW\ \DFB\ANIMBALL\0\27\ \DFB\ANIMBALL\238\27\ \DFB\ANIMBALL\14\27\ \DFB\ANIMBALL\224\27\ \DFB\ANIMBALL\28\27\ \DFB\ANIMBALL\210\27\ \DFB\ANIMBALL\42\27\ \DFB\ANIMBALL\196\27\ \DFB\ANIMBALL\56\27\ \DFB\ANIMBA If some of the kinetic energy of a system is lost, the collision is called inelastic. In an inelastic collision, some of the kinetic energy is lost as heat. While the total energy of the system is conserved, some of the kinetic energH\1\1\ \DTN\ When two objects collide, their velocities change. While their momentum must be conserved, their kinetic energy need not be.\ \DTN\ If the kinetic energy of a system is maintained, the collision is called elastic.\ \DTN\e vector sum of the momentums of the two balls after the collision is equal to that of the balls before the collision.\ \BESC\ \L\MOMENT VOCAB\ \RVR\4\ \BESC\ \L\MOMENT REVIEW\ \RRQ\4\ \BESC\ \L\COLLISIONS\ \SCH\2\2\ \DTCNW\Collisions\ \SCows the momentum vector of the first ball before the collision and the two thin lines show the momentum vectors of the two balls after the collision.\ \SWS\1\ \DCW\ \DF\MOMCONS5\1\0\1\0\ \SWT\\\R2\\1\ \DCW\ \DTN\ As in the previous example, thupward and the other downward after the collision.\ \DTN\ This shows the path of the moving ball before the collision, as well as the paths of both balls after the collision.\ \DF\MOMCONS4\1\1\1\0\ \SWT\\\R2\\1\ \DTN\ Here, the thick line shged.\ \DTN\ In this case, all of the momentum vectors are parallel. In other cases, they may not be.\ \DF\MOMCONS3\1\1\1\0\ \DTN\ This figure shows a moving ball striking a stationary ball off center.\ \DTN\ In this case, one ball moved m vectors of the two balls after they have separated. The dot represents the ball that has stopped. It has no arrow because its magnitude is zero and hence it is not moving in any direction.\ \DTN\ The total momentum of the system is unchancity to twice its original value.\ \SWS\1\ \DCW\ \SFF\0\\ \DF\MOMCONS1\1\1\1\0\ \SWT\\\6\\ \DTN\ This figure shows the momentum vectors of the two balls when they are moving together.\ \DFB\MOMCONS2\76\0\ \DTN\ This figure shows the momentuASE\238\27\279\37\ \SWT\\\7\\ \DTN\ In this example, two balls move along, connected to each other.\ \DTN\ An explosion between the two balls then pushes them apart.\ \DTN\ As a result, one ball stops moving and the other increases its velo\DFB\ANIMBALL\98\27\ \DFB\ANIMBALL\112\27\ \DFB\ANIMBALL\126\27\ \DFB\ANIMBALL\140\27\ \DFB\ANIMBALL\154\27\ \DFB\ANIMBALL\168\27\ \DFB\ANIMBALL\182\27\ \DFB\ANIMBALL\196\27\ \DFB\ANIMBALL\210\27\ \DFB\ANIMBALL\224\27\ \DFB\ANIMBALL\238\27\ \XER27\ \DFB\ANIMBALL\7\27\ \DFB\ANIMBALL\238\27\ \DFB\ANIMBALL\0\27\ \SWT\\\7\\ \DTNNW\Two 3 kg balls move together, each moving at 6 m/s, and then bounce back.\ \SWT\\\R1\\ \SQ\5\Y\ \SQT\Each ball has a momentum vector with a magnitude of ____ kgm/s.\ \SQF\N\5\ \SQH\p = mv\ \SQANS\18\ \SQLAST\No, each ball has a momentum vector with magnitude (3 kg) x (6 m/v) = 18 kgm/s\ \SQA\T\=\18\Right. Each ball has a momentum vector with magnitude (3 kg) x (6 m/v) = 18 kgm/s\ \DQ\ \BQESC\ \ALL\28\27\ \DFB\ANIMBALL\217\27\ \DFB\ANIMBALL\21\27\ \DFB\ANIMBALL\224\27\ \DFB\ANIMBALL\14\27\ \DFB\ANIMBALL\231\27\ \DFB\ANIMBALL\7\27\ \DFB\ANIMBALL\238\27\ \DFB\ANIMBALL\0\27\ \DCW\ \DTNNW\Two 3 kg balls move towards each other at velocitiFB\ANIMBALL\175\27\ \DFB\ANIMBALL\63\27\ \DFB\ANIMBALL\182\27\ \DFB\ANIMBALL\56\27\ \DFB\ANIMBALL\189\27\ \DFB\ANIMBALL\49\27\ \DFB\ANIMBALL\196\27\ \DFB\ANIMBALL\42\27\ \DFB\ANIMBALL\203\27\ \DFB\ANIMBALL\35\27\ \DFB\ANIMBALL\210\27\ \DFB\ANIMBL\140\27\ \DFB\POW2\105\0\ \DFB\ANIMBALL\140\27\ \DFB\ANIMBALL\98\27\ \DFB\ANIMBALL\147\27\ \DFB\ANIMBALL\91\27\ \DFB\ANIMBALL\154\27\ \DFB\ANIMBALL\84\27\ \DFB\ANIMBALL\161\27\ \DFB\ANIMBALL\77\27\ \DFB\ANIMBALL\168\27\ \DFB\ANIMBALL\70\27\ \DB\ANIMBALL\28\27\ \DFB\ANIMBALL\210\27\ \DFB\ANIMBALL\42\27\ \DFB\ANIMBALL\196\27\ \DFB\ANIMBALL\56\27\ \DFB\ANIMBALL\182\27\ \DFB\ANIMBALL\70\27\ \DFB\ANIMBALL\168\27\ \DFB\ANIMBALL\84\27\ \DFB\ANIMBALL\154\27\ \DFB\ANIMBALL\98\27\ \DFB\ANIMBALalls are irrelevant.\ \SQA\F\=\0\No, energy is not a vector. When you add energies, you simply add their magnitudes.\ \DQ\ \BQESC\ \SQSEND\ \SWS\1\ \DCW\ \DFB\ANIMBALL\0\27\ \DFB\ANIMBALL\238\27\ \DFB\ANIMBALL\14\27\ \DFB\ANIMBALL\224\27\ \DFor, it is just a magnitude.\ \SQANS\108\ \SQLAST\No, the total kinetic energy is the sum of the kinetic energies of the parts of the system: 54 J + 54 J = 108 J.\ \SQA\T\=\108\Right. The kinetic energy is not a vector, so the directions of the b E = 1/2 mv2\ \SQA\F\=\9\No, E = 1/2 mv2\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \SQF\N\5\ \SQT\Given that the kinetic energy of each ball is 54 J, the total kinetic energy of the system is ____ J.\ \SQH\Remember that energy is not a vect/2 mv2\ \SQANS\54\ \SQLAST\No, the energy = 1/2 mv2 = 1/2 (3 kg) x (6 m/s)2 = 54 J\ \SQA\T\=\54\Right. The energy = 1/2 mv2 = 1/2 (3 kg) x (6 m/s)2 = 54 J\ \SQA\F\=\108\No, E = 1/2 mv2\ \SQA\F\=\18\No,te directions!\ \SQA\T\=\0\Right. The two vectors cancel each other out.\ \SQA\T\=\0 kgm/s\Right. The two vectors cancel each other out.\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \SQF\N\5\ \SQT\Each ball has ____ J of kinetic energy .\ \SQH\E = 1momentum vectors together.\ \SQANS\zero\ \SQLAST\No, since the two vectors are equal and opposite, their sum is zero.\ \SQA\T\CONTAINS\zero\Right. The two vectors cancel each other out.\ \SQA\F\CONTAINS\36\No, the two momentums are in opposio it's moving toward the left.\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \SQF\S\8\ \SQT\The momentums of the two balls are 18 kgm/s moving toward the right and 18 kgm/s moving toward the left. What is the total momentum of the system?\ \SQH\Add the two right?\ \SQF\S\6\ \SQH\Is the ball moving toward the LEFT or toward the RIGHT?\ \SQANS\left\ \SQLAST\No, the ball coming from the right is moving toward the left.\ \SQA\T\CONTAINS\left\ \SQA\T\CONTAINS\right\No, it's coming from the right, sthe right.\ \SQA\T\CONTAINS\right\ \SQA\T\CONTAINS\left\No, it's coming from the left, so it's moving toward the right.\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \SQF\S\6\ \SQT\What is the direction of the momentum vector for the ball coming from the DCW\ \SQ\ \SQT\What is the direction of the momentum vector for the ball coming from the left?\ \SQF\S\6\ \SQH\Is the ball moving toward the LEFT or toward the RIGHT?\ \SQANS\right\ \SQLAST\No, the ball coming from the left is moving toward es of 6 m/s, one coming from the left and one from the right.\ \DTN\After colliding, they bounce back apart, each moving at a velocity of 4 m/s.\ \SWT\\\R1\\ \SQ\5\Y\ \SQF\N\5\ \SQT\The magnitude of the momentum vector of each ball as it bounces back is ____ kgm/s.\ \SQH\p = mv\ \SQANS\12\ \SQLAST\No, each ball has a momentum vector with magnitude (3 kg) x (4 m/v) = 12 kgm/s\ \SQA\T\=\12\Right. Each ball has a momentum vector with magnitude (3 kg) x (4 m/v) = 12 kgm/s\ \DQ\ \BQESvxy forces act on the system.\ \L\displacement\ \2\a change of position in a particular direction.\ \L\elastic collision\ \5\a collision in which the total kinetic energy of the system remains constant.\ \L\energy\ \2\the capacity to do work or toTue Sep 4 16:24:12 1990 \L\conservation of energy and matter\ \2\the sum of the matter and energy in a closed system is constant.\ \L\conservation of momentum\ \4\the total vector momentum of a system will remain constant, provided that no external____ J.\ \SQH\Remember that energy is not a vector, it is just a magnitude.\ \SQANS\48\ \SQLAST\No, kinetic energy is not a vector, so the directions of the balls are irrelevant and the total kinetic energy is just the sum of the individual energi48\No, E = 1/2 mv2\ \SQA\F\=\12\No, E = 1/2 mv2\ \SQA\F\=\6\No, E = 1/2 mv2\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \SQF\N\5\ \SQT\Given that the kinetic energy of each ball is 24 J, the total kinetic energy of the system is now as ____ J of kinetic energy.\ \SQH\E = 1/2 mv2\ \SQANS\24\ \SQLAST\No, the energy = 1/2 mv2 = 1/2 (3 kg) x (4 m/s)2 = 24 J\ \SQA\T\=\24\Right. The energy = 1/2 mv2 = 1/2 (3 kg) x (4 m/s)2 = 24 J\ \SQA\F\=\pposite, their sum is zero.\ \SQA\T\IS CONTAINED IN\zero~0 kgm/s\Right. The two vectors cancel each other out.\ \SQA\F\CONTAINS\24\No, the two momentums are in opposite directions!\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \SQF\N\5\ \SQT\Each ball now hVIEWXPOWERYPOWER INTRO^POWER CALC/vPOWER VOCABRvPOWER REVIEWqvMOMENTUMwMOMENT INTROMOMENT IMPULSE@MOMENT CONSERVEMOMENT VOCABMOMENT REVIEWܖCOLLISIONSVOCABREVIEWûMORE}MAINMENUNO MAINMENUMOMENTUM & WORK>WORKsWORK INTRO\WORK VECTORWORK VARY(WORK VOCAB(WORK REVIEW)ENERGY*ENERGY INTRO13ENERGY CONSERVEO;ENERGY POTENTIAL]MENERGY KINETICWENERGY VOCABWENERGY REof the two balls are 12 kgm/s moving toward the left and 12 kgm/s moving toward the right. What is the total momentum of the system?\ \SQH\Add the two momentum vectors together.\ \SQANS\zero\ \SQLAST\No, since the two vectors are equal and oced back and is moving back toward the right.\ \SQA\T\CONTAINS\right\ \SQA\T\CONTAINS\right\No, the ball coming from the right has bounced back and is moving back toward the right.\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \SQF\S\8\ \SQT\The momentums the collision, what is the direction of the momentum vector for the ball initially coming from the right?\ \SQF\S\6\ \SQH\Is the ball moving toward the LEFT or toward the RIGHT?\ \SQANS\right\ \SQLAST\No, the ball coming from the right has bounng from the left has bounced back and is moving back toward the left.\ \SQA\T\CONTAINS\left\ \SQA\T\CONTAINS\left\No, the ball coming from the left has bounced back and is moving back toward the left.\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \SQT\After C\ \DCW\ \SQ\ \SQT\After the collision, what is the direction of the momentum vector for the ball initially coming from the left?\ \SQF\S\6\ \SQH\Is the ball moving toward the LEFT or toward the RIGHT?\ \SQANS\left\ \SQLAST\No, the ball comi produce heat.\ \L\gravitational potential energy\ \2\the energy an object gains as a result of being moved against a gravitational force.\ \L\horsepower\ \3\a unit of power equal to approximately 746 watts.\ \L\inelastic collision\ \5\a collisTN\ Disk #1, Force and Motion, describes the basic concept of vectors in the section Friction and Vectors. Disk #4, Waves and Sound, describes another form of acceleration in the section on Simple Harmonic Motion. Disk #3, Heat, describesc energy.\ \SQA\F\=\108\No, this is the amount that was initially present. Not all of it was lost.\ \DQ\ \SQSEND\ \BESC\ \L\VOCAB\ \RVR\0\ \BESC\ \L\REVIEW\ \RRQ\0\ \BESC\ \L\MORE\ \DCW\ \SCH\2\2\ \DTCNW\More On This Topic\ \SCH\1\1\ \Dafter the collision, so 60 J were converted into heat.\ \SQA\T\=\60\Right. Only 48 of the initial 108 J of kinetic energy remained after the collision, so 60 J were converted into heat.\ \SQA\F\=\48\No, this is the amount that remained as kinetiSQF\N\5\ \SQT\How much of the initial kinetic energy was converted into heat during the collision?\ \SQH\The initial content was 108 J and the final content was 48 J.\ \SQANS\60\ \SQLAST\No, only 48 of the initial 108 J of kinetic energy remained es, or 48 J.\ \SQA\T\=\48\Right. The kinetic energy is not a vector, so the directions of the balls are irrelevant.\ \SQA\F\=\0\No, energy is not a vector. When you add energies, you simply add their magnitudes.\ \DQ\ \BQESC\ \DCW\ \SQ\5\Y\ \pdisk2   Q w"1u92?@%C#E7GSdTU0]]_KgRi'kmoq$t|‚9Sز״al energypowerwattWwork @conservation of energy and matterconservation of momentum.displacement|elastic collisionenergy4gravitational potential energyhorsepowerinelastic collision^joulekinetic energymomentumjpotenti transferred.\ \L\watt\ \3\the basic unit of power - 1 joule per second.\ \L\work\ \1\a measurement of energy expended, calculated by multiplying the displacement of an object by the force applied in the direction of the displacement.\ eleration.\ \L\momentum\ \4\the product of a moving object's mass and velocity.\ \L\potential energy\ \2\the energy an object gains as a result of being moved against a force.\ \L\power\ \3\the rate at which work is done or energy is emitted orion in which some of the system's kinetic energy is converted into heat energy.\ \L\joule\ \1\the basic unit of work - product of 1 newton acting through a distance of 1 meter.\ \L\kinetic energy\ \2\the energy an object gains as the result of acc another form of energy, in the section on Heat and Energy.\ \BESC\ pcreds.iec p2creds.iec smallball.bits twoballs.iec twoballs2.iec work0.iec work1.iec work2.iec work3.iec vary1.iec vary2.iec vary3.iec animball.iec pow1.iec pow2.iec momcons1.iec momcons2.iec momcons3.iec momcons4.iec momcons5.iec small<`** pUU x@**<`UUp **x@ UU<`**pUUx@** ( 7|x} " `pp8X8pp`?}ny? ** x@ UU<` **p UUx@ **<` UUp ** 1ox@x UU **# UU#( ** (!)@`0 **@`0 UU@`0  ** @`0 UUp@`0 **?m UUp**x@UUball.iec