Chapter 8 Potential Energy and Conservation of Energy Important Terms (For chapters 7 and 8)

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1 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Chapter 8 Potental Energy and Conservaton o Energy Important Terms (For chapters 7 and 8) conservatve orce: a orce whch does wor on an object whch s ndependent o the path taen by the object between ts startng pont and ts endng pont conserved propertes: any propertes whch reman constant durng a process energy: the non-materal quantty whch s the ablty to do wor on a system joule: the SI unt or energy, and wor, equal to one Newton-meter netc energy: the energy a mass has by vrtue o ts moton law o conservaton o energy: the total energy o a system remans constant durng a process mechancal energy: the sum o the potental and netc energes n a system potental energy: the energy an object has because o ts poston power: the rate at whch wor s done or energy s dsspated watt: the SI unt or power equal to one joule o energy per second wor: the scalar product o orce and dsplacement W Fs F cos s W Fds Fds KE W K KE KE mv mv PE mgh E E E 0 PE KE PE KE mv U W mg h h W g g mech 0 E mech W Fs s P F F v F cos v t t t P m a v F cos v where W = wor F = orce s = dsplacement F. s = scalar product o orce and dsplacement KE = netc energy v = velocty or speed m = mass PE = potental energy (denoted as U ) g = acceleraton due to gravty h = heght above some reerence pont P = power t = tme = ntal state = nal state

2 Pro. Dr. I. Nasser Chapter8_I November 3, POTENTIAL ENERGY (Done n Chapter 7) H.W. Use the orce to calculate the wor done, through the path AC and the path ABC. Ans (30 J). What s your comment? F ext 0 ˆj H.W. What s the wor done, W G H.W. What s the relaton between, by the gravtatonal eld? W ext and W ext W G? Dene the P.E. Dene: ext = external orce, G = gravtaton eld W U ext G W W G ext 0 In summary: The wor done could be calculated n two derent ways: F ext. d (I) Wext K mv mv (II) In rsng (lowerng) a mass m to (rom) a heght h, the wor done by the gravtatonal orce s, U G. W G W W G G mgh mgh Compare ths wth the change n the gravtatonal potental energy: U U G G mgh mgh Ths mples: U W G G Ths s also appled or the sprng: U W x x S S

3 Pro. Dr. I. Nasser Chapter8_I November 3, CONSERVATION OF MECHANICAL ENERGY Conservatve orce A orce s conservatve the wor done by ths orce when actng on a movng object between two ponts s ndependent o the path the partcle taes between these two ponts. The wor done by conservatve orce depends only on the coordnates o the ntal and nal ponts and ndependent o the path taen between these two ponts. (see the ollowng gure). Let be the wor done n movng rom pont a to pont b W ab () W ab () through the path (), and be the wor done n movng rom pont a to pont b through the path (). I the orce s conservatve then Wab () Wab () The wor done by conservatve orce n a reversed path equal the negatve o the wor done by ths orce n the non-reversed path. W () W () W () W ab ba ba ab W ab () W () () W () 0 Snce the path aba s a closed path then, we can say that: ba the wor done by conservatve orce when the object moves n a closed path s zero. The wor done by gravtatonal orce - When a body o mass m moves rom a pont o heght h to a pont o heght h rom the earth surace, under the eect o gravty, the wor done by the gravty s gven by W mg( h h ) ba G It s negatve because the moton s aganst the weght mg. - I the body maes a round trp and returns to ts ntal heght, h h, then the wor done by gravty s zero. Snce the wor done by any orce equal the change n netc energy, then n ths case the change n netc energy s zero. The nal netc energy equal the ntal netc energy and the nal speed equal the ntal speed. K K v Nonconservatve orce A orce s non-conservatve the wor done by ths orce on a partcle movng between two ponts depends on the path taen e.g. rcton orces. The property o non-conservatve orce s that v Wab () Wab () W () W () 0 ab ba 3

4 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Potental energy and the wor done by conservatve orce The wor done by the conservatve orce equal the decrease n potental energy. I the wor s, then Wc U.e. the wor done by the conservatve orce equal the negatve o the change n potental energy. W c Conservaton o mechancal energy When a conservatve orce acts on a movng object, then rom the wor energy theorem () W c W K c K Where s the wor done and s the change n netc energy. Snce the orce s conservatve then Wc U () s the change n potental energy. From () and () we get. K U E s the total energy. K K U K U 0 ( KU) 0 K s the nal netc energy, and s the ntal netc energy, U U s the nal potental energy K U K U E E s the ntal potental energy, The law o conservaton o mechancal energy Ths law states that the total mechancal energy o a system remans constant the only orce actng on the body s conservatve orce. I the potental energy decreases (or ncreases) the netc energy ncreases (or decreases) by the same amount Example: Use conservaton o total mechancal energy to compute the nal velocty o an object dropped rom some heght, h, just beore t stres the ground. Here, only conservatve orces are present. Consequently, we have: E E K U K U Dene the zero potental energy reerence level at the nal poston conservaton o mechancal energy proceeds as ollows: At pont, K 0, U mgh E K U mgh m m At pont, K v, U 0 E K U v Use the conservaton o mechancal energy proceeds as E 0 m v Whch s the same as the nematc equaton. E E mgh 0 v gh, see gure. The 4

5 Pro. Dr. I. Nasser Chapter8_I November 3, Example: To what heght wll a ball rse tossed upwards wth an ntal velocty o 0 m/s? Agan, only conservatve orces are present: E E K U K U or mv mgh mv mgh. I we dene the ground as beng at heght, (zero potental energy reerence level), then h h h 0 and conservaton o mechancal energy proceeds as ollows: At pont, K mv, U 0 E K U mv At pont, K 0, U mgh E K U mgh v mv 0 0 mgh h 5. m. g Note: We could also use Wor-Energy theorem: W K mv v mg. h 0 mv gh h 5. m Example: A ball sldes wthout rcton around a loop-the-loop (see the gure). A ball s released, rom rest, at a heght h rom the let sde o the loop o radus R. What s the speed o the ball at the hghest pont o the loop? (g = acceleraton due to gravty) Apply the conservaton o the mechancal energy at pont I and at the hghest pont o the loop, we nd: At pont At pont, K 0, U mgh E K U mgh m m, K v, U mg( R) E K U v mgr Applyng the conservaton o mechancal energy: m v mgr mgh v E 0 E E 0 g( h R) 5

6 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Example: A projectle s red rom the top o a 40 m hgh buldng wth a speed o 0 m/s. What wll be ts speed when t stres the ground? E E mv mgh mv v v gh m/s Example: A 0 g bloc sldes down a smooth surace. It s released rom rest at the top o an nclne (38 o to the horzontal) at a heght 0 meters. What s ts speed at the bottom o the nclne? We ve seen ths problem beore. Ths method o solvng the problem yelds a soluton but t depends upon resolvng vectors, summng orces, determnng acceleraton, and nally nematcs! Now consder usng conservaton o mechancal energy to address the same problem: Snce only conservatve orces are present: K U K U I we dene the ground as beng at heght h 0 h h At pont, K 0, U mgh E K U mgh At pont,, (zero potental energy reerence level), then mv, 0 mv K U E K U Use the conservaton o mechancal energy proceeds as E 0 m v E E mgh 0 v gh Then mv mgh mv mgh. I we dene the ground as beng at heght h 0, (zero potental energy reerence level), then h h mv 0 0 mgh v gh 4. m/s. Ths s a much more compact and elegant method o soluton

7 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Conservaton o Mechancal Energy wth non-conservatve orces When non-conservatve orces are present, total mechancal energy s not conserved,.e., the total energy o the system changes ( ). The only non-conservatve orce we wll deal wth s rcton. E E Frctonal Forces Frctonal orces result n energy beng dsspated rom systems. Frctonal orces reduce the netc energy o a system by some amount that generally s easy to compute. The result s that the ntal and nal energes o a non-conservatve system are not the same. The energes, n act, der by the amount o wor done by the non-conservatve orce (rcton): W W d KE or W d E.. nc nc Example: A.0 g bloc s ntally movng to the rght on a rough horzontal surace, wth Calculate ts speed ater movng.0 m. K 0.5, at a speed o 5.0 m/s. At state : At state : K mv ()(5), U 0, E 5 J. K mv v, U 0, E v.. d mgd ()(9.8)() (I) Wnc E E v 5 (II) Equatng (I) and (II), we get v m/s Example: A 0 g bloc sldes down a rough surace (µ =0.). It s released rom rest at the top o an nclne (38 o to the horzontal) at a heght 0 meters. What s ts speed at the bottom o the nclne? Usng conservaton o energy. Snce a non-conservatve orce s present W d KE or W d E. Usng the latter:... d mgd cos Wnc mv mv mgh mgh mv mgh Equatng (I) and (II), we get v gh mgd cos. m/s (I) (II) 7

8 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Example: A ser starts rom rest at the top o a 60 m tall hll. The porton o the hll covered wth snow s essentally rctonless but lat secton at the base o the hll covered wth mud. I the ser weghs 80 g, and the coecent o netc rcton between rental ss and mud s 0.8, how much dstance s requred or the ser to come to rest? A non-conservatve orce, rcton, s present here,, and total mechancal energy s not conserved. E E 0 E E mgh (80 g)(9.8 m/s )(60 m) 5440 J s. s mgs (I) Wnc E 5440 J (II) Equatng (I) and (II), we get 5440 J s 00 m (80 g)(9.8 m/s )(0.8) Example: Ths tme our 80 g ser starts rom rest at the top o a 60 meter tall hll. The downhll porton o the run s snow covered and essentally rctonless but a lat secton at the base o the hll has a bare spot meters n length. The ser ss straght down the hll, over the bare spot at the bottom, hts the counterweght on the bullwheel compressng the dampng sprng 50 centmeters n the process. I the sprng constant s 0 6 N/m, what s the coecent o netc rcton between rental ss and bluegrass? A non-conservatve orce, rcton, s present here, E E, and total mechancal energy s not conserved. E E (80 g)(9.8 m/s )(60 m) 5440 J 6 x mgh (0.5)(0 N/m)(0.5 m) 5000 J W E 440 J. s mgs s 440 J 0.8 (80 g)( m)(9.8 m/s ) 8

9 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Extra Problems Example: A ball sldes wthout rcton around a loop-the-loop (see the gure). A ball s released, rom rest, at a heght h rom the let sde o the loop o radus R. What s the rato (h/r) so that the ball has a speed at the hghest pont o the loop? (g = acceleraton due to gravty) Apply the conservaton o the mechancal energy at the begnnng o the trp and at the hghest pont o the loop, we nd: At pont, K 0, U mgh E K U mgh v Rg 5 v ( ), ( ) m m Rg mgr At pont, K U mg R E K U 5 h 5 E 0 E E mgr mgh 0 R Example: A worer does 500 J o wor n movng a 0 g box a dstance D on a rough horzontal loor. The box starts rom rest and ts nal velocty ater movng the dstance D s 4.0 m/s. Fnd the wor done by the rcton between the box and the loor n movng the dstance D. Total wor by the man s used to change the K.E. o the body and to overcome the rcton. So, K W W man W K W mv J Example: A.0 g bloc s released rom rest 60 m above the ground. Tae the gravtatonal potental energy o the bloc to be zero at the ground. At what heght above the ground s the netc energy o the bloc equal to hal ts gravtatonal potental energy? (Ignore ar resstance) Apply the conservaton o the mechancal energy at the begnnng o the trp and at the x-poston, we nd: K U K U x x 0 mgh mgx mgx x h 40 m Example: A.-g bloc starts rom rest on a rough nclned plane that maes an angle o 5 wth the horzontal. The coecent o netc rcton s 0.5. As the bloc goes.0 m down the plane, nd the change n the mechancal energy o the bloc. E W Nd mg cos30 d nc man J

10 Pro. Dr. I. Nasser Chapter8_I November 3, Example: A 0.50 g bloc attached to an deal sprng wth a sprng constant o 80 N/m oscllates on a horzontal rctonless surace. The speed o the bloc s 0.50 m/s, when the sprng s stretched by 4.0 cm. The maxmum speed the bloc can have s: (Ans: 0.7 m/s) mv mv x v m/s Example: A 0.0 g bloc s released rom rest at 00 m above the ground. When t has allen 50 m, ts netc energy s: K U mgh J. g Example: A 4.0 g bloc s ntally movng to the rght on a horzontal rctonless surace at a speed o 5.0 m/s. It then compresses a horzontal sprng o sprng constant 00 N/m. At the nstant when the netc energy o the bloc s equal to the potental energy o the sprng, the mechancal energy o the bloc-sprng system s: E 0 E E (or at any place) E mv J Example: A 5.0 g bloc starts up a 30 nclne wth 98 J o netc energy. The bloc sldes up the nclne and stops ater travelng 4.0 m. The wor done by the orce o rcton between the bloc and the nclne s: W E K U ( K K ) ( U U ) (0 98) ( sn 30 ) 00 J Example: A.0 g object s connected to one end o an unstretched sprng whch s attached to the celng by the other end and then the object s allowed to drop. The sprng constant o the sprng s 96 N/m. How ar does t drop beore comng to rest momentarly? (Ans: 0.0 m) Apply the conservaton o the mechancal energy: mg E x mgh h m. 0

11 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Example: An deal sprng (compressed by 7.00 cm and ntally at rest,) res a 5.0 g bloc horzontally across a rctonless table top. The sprng has a sprng constant o 0.0 N/m. The speed o the bloc as t leaves the sprng s: Apply the conservaton o the mechancal energy: E 0 K K U U 0 G G ( mv 0) 0 x 0 0 v x (0.07).56 m/s. m Example: A small object o mass m on the end o a massless rod o length L s held vertcally, ntally. The rod s pvoted at the other end O. The object s then released rom rest and allowed to A small object o mass m on the end o a massless rod o length L s held vertcally, ntally. The rod s pvoted at the other end O. The object s then released rom rest and allowed to swng down n a crcular path as shown n the gure. What s the speed (v) o the object at the lowest pont o ts swng? (Assume no rcton at the pvot) Apply the conservaton o the mechancal energy at the begnnng o the trp and at the lowest pont o the moton, we nd: E 0 K K U U 0 ( mv 0) mg L 0 0 v 4Lg G G Example: A projectle s red rom the top o a 40 m hgh buldng wth a speed o 0 m/s. What wll be ts speed when t stres the ground? E E mv mgh mv v v gh m/s Example:A projectle o mass 0.0 g s red wth an ntal speed o 0 m/s at an angle o 60 degrees above the horzontal. The netc energy o the projectle at ts hghest pont s:

12 Pro. Dr. I. Nasser Chapter8_I November 3, 07 v v gd ; y y K mv J 0 0sn 60 gd d 5.3 m v v gd Example: A 5.0-g object s pulled along a rough horzontal surace at constant speed by 5 N orce actng above the horzontal (see Fgure). (a) How much wor s done by the rcton orce as the object moves 6.0 m? (b) What s the ncrease n thermal energy o the bloc loor system? (a) constant speed a K 0 K Total wor done W W 0 F W W F cos30 d = J F 30 (b) The ncrease n thermal energy wll be Eth W 77.9 J Example: A 0.6-g ball s suspended rom the celng at the end o a.0-m strng. As ths ball swngs, t has a speed o 4.0 m/s at the lowest pont o ts path. What maxmum angle does the strng mae wth the vertcal as the ball swngs? Apply the conservaton o the mechancal energy we nd: v mv mgh h 0.86 m g Wth h h cos cos cos Example: A.0 g bloc s pulled at a constant speed o. m/s across a horzontal rough surace by an appled orce o N drected 30 above the horzontal. At what rate s the rctonal orce dong wor on the bloc? The acceleraton here s zero, snce we have constant velocty. The wor done by the appled orce s used to overcome the rcton wor, so: K W W 0 W W F cos30 d F F dw F P Fcos30 v W dt

13 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Example: A 0 g object s dropped vertcally rom rest. Ater allng a dstance o 50 m, t has a speed o 6 m/s. How much wor s done by the ar resstance on the object durng ths descent? ) ( ) W E E (0 m 6 mg J nc Example: As a partcle moves rom pont A to pont B only two orces act on t: one orce s non-conservatve and does wor = -30 J, the other orce s conservatve and does +50 J wor. The change o the netc energy o the partcle s: K W W J nc c Q: A 6 g crate alls rom rest rom a heght o.0 m onto a sprng scale wth a sprng constant o N/m. Fnd the maxmum dstance the sprng s compressed. I the sprng s compressed a dstance x, we can apply the conservaton o mechancal energy; E E E K K U U G G U U S S 0; mg ( x ) 0 0 x 0 Solve o x, wll nd: x 0.4 m; We too the nal poston o the sprng as a reerence o our potental energy Q4. A bloc s sldng down on a rough nclned plane. A man apples a orce to reduce the acceleraton o the bloc. Let W be the wor done by the rcton orce, Wm the wor done by the man, and Wg the wor done by the gravtatonal orce. Whle the bloc s sldng down, whch o the ollowng s TRUE? Ans: A A) W < 0, Wm < 0, Wg > 0 B) W < 0, Wm > 0, Wg < 0 C) W < 0, Wm < 0, Wg < 0 D) W < 0, Wm > 0, Wg > 0 E) W > 0, Wm > 0, Wg > 0 3

14 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Q5. A ball s launched upward rom the edge o a cl. Whch o the graphs shown n Fgure 3 could possbly represent how the netc energy o the ball changes durng ts lght? A) a B) b C) c D) d E) e K K (a) t K K (b) Fgure 3 t Ans: A (c) K t (d) t t (e) Q6: A man pushes a bloc up an nclne at a constant speed. As the bloc moves up the nclne, only one statement s true: A) Its netc energy remans constant and the potental energy o bloc-earth system ncreases B) Its netc energy and the potental energy o bloc-earth system both ncrease C) Its netc energy ncreases and the potental energy o bloc-earth system remans constant D) Its netc energy decreases and the potental energy o bloc-earth system remans constant E) Its netc energy decreases and the potental energy o bloc-earth system decreases Ans: Q8: A 80 g ser starts rom rest at heght H = 5 m above the end o a rctonless s-jump ramp and leaves the ramp at angle = 30, as shown n Fgure 4. What s the maxmum heght h o hs jump above the end o the ramp? Neglect the eects o ar resstance. Ans: A) 6.3 m B) 5.6 m C) 7. m D) 4.3 m E) 8. m Fgure

15 Pro. Dr. I. Nasser Chapter8_I November 3, 07 Q6: A.00 g ball s thrown wth an ntal velocty o v 8ˆ 0 ˆ o j, where vo s n m/s. What s the maxmum change n the potental energy o the ball-earth system durng ts lght? Ans: Q8: The only orce actng on a partcle s a conservatve orce F. I the partcle s at pont A, the potental energy o the system assocated wth F and the partcle s 80 J. I the partcle moves rom pont A to pont B, the wor done on the partcle by s + 0 J. What s the potental energy o the system wth the partcle at B? Ans: Example: An 80 g ser starts rom rest at the top o a 60 meter tall hll and attans a velocty o 36 m/s by the tme they reach the bottom o the 00 meter run. What s the coecent o netc rcton between ther ss and the snow? A non-conservatve orce, rcton, s present here, ( E E ), and total mechancal energy s not conserved. E K U 0 mgh (80 g)(9.8 m/s )(60 m) 5440 J E K U mv W E J. s mhs (80 g)(9.8 m/s )(00 m) 0 (0.5)(80 g)(36 m/s ) 5840 J J 0.47 s F 5

16 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 WORK DONE BY A SPRING FORCE Chapter 8 Revew Problems (Sprng) - We are dealng wth massless sprng, whch mples - Hooe's "Law": Fsprng x Ksprng 0 3- The wor done by an external orce F ext (such as the orce rom my hand whch s opposte to F sprng) to stretch or compress a sprng by an amount x s gven by x x x ext Fext dx xdx x x x x x x W Ths s the general expresson or wor done by an external orce Fext that can be used n all cases, we now x and x. Note that: Startng rom equlbrum poston,.e. x 0, then the above equaton reduces to W ext x. Ths condton mples: - The wor done by the external orce s always postve, regardless o whether the sprng s stretched ( x postve) or compressed ( - The expresson Feld ext x negatve). W W means that the wor done by the sprng s always negatve, regardless o whether the sprng s stretched ( postve) or compressed ( negatve). U S WS Wext x stored n the sprng s always postve, regardless o whether the sprng s stretched means that the potental energy - The expresson x x ( postve) or compressed ( negatve) Q: A g bloc, restng on a rctonless surace s pushed 8.00 cm nto a lght sprng, N/m. It s then released. What s the velocty o the bloc as t just leaves the sprng? We can solve ths by analyzng the energy stuaton. There are only two terms we need worry about, the potental energy stored n the sprng and the netc energy when the sprng uncols and the bloc s released., m, s state : K mv 0, U x E x, m, s state : K mv 0, U 0 E mv 0 Conservaton o mechancal energy mples: E E x mv g m m v x m =.6 m g s s x x

17 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 Q: A 55 g bloc s travelng along a smooth surace wth a velocty o.5 m/s. It runs head on nto a sprng ( 35 N/m ). How ar s the sprng compressed? m m 0.55 g mv x x v.5 = m s g m 35 m s Q: In the gure, a bloc (M =.0 g) sldes on a rctonless horzontal surace towards a sprng wth a sprng constant 000 N/m. The speed o the bloc just beore t hts the sprng s 6.0 m/s. How ast s the bloc movng at the nstant the sprng has been compressed 5 cm? state : K, m mv, U, s 0 E mv (6 ), m, s state : K mv, U x E mv x Conservaton o mechancal energy mples: E E 0 v 6 x 0 v v 3.67 m/s Q: A bloc o mass m=3.0 g s ept at rest ater t has compressed a horzontal massless sprng ( 500 N/m ) by 0.5 m, as shown n the gure. When the bloc s released, t travels a dstance S on a horzontal rough surace ( 0.4 ) beore stoppng. Calculate the dstance S., m state : K mv 0, U x E x (500)(0.5) 5.65, m state : K mv 0, U 0 E 0 0. d mgs 0.4(3)(9.8)(S) (I) Wnc E E 5.65 (II) Usng the equaton: E E Wnc and equatng (I) and (II), we get 5.65 s m 0.4(3)(9.8)

18 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 Q: A sprng-loaded gun res a dart at an angle rom the horzontal. The dart gun has a sprng wth sprng constant that compresses a dstance x. Assume no ar resstance. What s the speed o the dart when t s at a heght h above the ntal poston? E E K U U K U U g, elas, g, elas, 0 0 x mv +mgh 0 v x g h m Notce that the angle never entered nto the soluton Q: A 5 g box s pushed along a horzontal smooth surace wth a orce o 35 N or.5 m. It then sldes nto a sprng 500 N/m and compresses t. Fnd the dstance that the sprng s compressed. Fnd the acceleraton: F g m m F ma a 35 =.4 m s 5 g s Fnd the nal velocty : v v0 ax v ax.4.5 m.05 m/s Now we can nd the netc energy o the thng beore compressng: m Ko mv g J s We can now nd the dsplacement o the sprng usng the netc energy we just ound: K 5.5 N m 0 K0 x x 500 N/m = 0.6 m Example: A bloc o mass m = 0 g s connected to un-stretched sprng 400 N/m (see Fgure). The bloc s released rom rest. I the pulley s mass less and rctonless, what s the maxmum extenson o the sprng? Tae our reerence pont or the gravtatonal potental energy at the ntal poston o the bloc, and apply the conservaton o the mechancal energy usng the maxmum extenson o the sprng s A, one can nds: E E A mga 0 0 mg A 0.49 m

19 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 Example: A bloc o mass m =.0 g s dropped rom heght h = 40 cm onto a sprng o sprng constant 960 N/m (see Fgure). Fnd the maxmum dstance, s, the sprng s compressed. We denote m as the mass o the bloc, h = 0.40 m as the heght rom whch t dropped (measured rom the relaxed poston o the sprng), and s the compresson o the sprng (measured downward so that t yelds a postve value). Our reerence pont or the gravtatonal potental energy s the ntal poston o the bloc. At the ntal state, K K K 0 0, U U U 0 0 E 0 0 At the nal state, K,mass,sprng,mass,sprng,mass,sprng,mass,sprng K K 0 0, U U U mg( h s) s E mg( h s) s The bloc drops a total dstance h s, and the nal gravtatonal potental energy s mg( h s). The sprng potental energy s s n the nal stuaton, and the netc energy s zero both at the begnnng and end. Snce energy s conserved E E E 0 0 mg( h s) s whch s a second degree equaton n x. Usng the quadratc ormula, ts soluton s mg mg mgh s Now mg = 9.6 N, h = 0.40 m, and 960 N m, and we choose the postve root so that s > s 0.0 m

20 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 Q: A sprng and a bloc are n the arrangement o Fgure. When the bloc s pulled out to x = 4.00 cm, we must apply a orce o magntude 360 N to hold t there. The bloc s pulled to x =.0 cm and then released. How much wor does the sprng do on the bloc as the bloc moves rom x = 5.00 cm to x = 3.00 cm? Frst use Hoo s law to calculate the constant : F 360 F x 9000 N/m; x 40 4 W x x J s Example: A.00 g pacage s released on a rough 53. o nclne at d = 4.00 m rom a long sprng o orce constant 0 N/m. The sprng s attached to the bottom o the nclne as shown n Fgure. I the maxmum compresson o the sprng s x =.00 m, what s the wor done by the rcton orce? Tae our reerence pont or the gravtatonal potental energy at the ntal poston o the bloc, and apply the rule: K 0, U U U x mg( h) 0 W 9.8 ( d x )sn(53. ) 8.4 J Example: A.0 g ball s launched rom a sprng 35 N/m that has been compressed a dstance o 5 cm. The ball s launched horzontally by the sprng, whch s.0 m above the dec. (a) What s the velocty o the ball just ater t leaves the sprng? (b) What s the horzontal dstance the ball travels beore t hts? (c) What s the netc energy o the ball just beore t hts? (a) x mv v x m m.0 g.9 m/s (b) Fnd the tme to all: y.0 y at t a 9.8 D vt m s (c) To drectly calculate the netc energy o the ball, we would have to calculate what t s velocty s just beore t hts. Ths would be a complcated problem vectors, x and y components etc. Much easer to calculate t usng conservaton o energy. Its energy at the top, whch wll be the potental energy n the sprng plus the gravtatonal potental energy because o the ball s heght, has to equal ts netc energy at the bottom. We could also use 5

21 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 ts netc energy plus ts gravtatonal potental energy at the top just beore t leaves the table (ths s because ts netc energy must equal the potental energy stored n the sprng beore launch). mgy 0 x0 K nal K nal mgy0 x J Example: In the gure, A.00 g bloc stuated on a rough nclne, wth, s connected to a sprng o neglgble mass and sprng constant 00 N/m. The bloc s released rom rest when the sprng s un-stretched. I the pulley s massless and rctonless and the bloc moves 0.0 cm down the nclne beore comng to rest, then nd the coecent o netc rcton between the bloc and nclne. use the relaton K U U W () 37.0 g s nc wth the condtons: K K K 0, U mgd sn 37.36, g U x g Wnc mgd cos From () to (), we nd: Example: A.0 g object s connected to one end o an unstretched sprng whch s attached to the celng by the other end and then the object s allowed to drop. The sprng constant o the sprng s 96 N/m. How ar does t drop beore comng to rest momentarly? Apply the conservaton o the mechancal energy: mg E 0 x mgh h 0. m Example: An deal sprng (compressed by 7.00 cm and ntally at rest,) res a 5.0 g bloc horzontally across a rctonless table top. The sprng has a sprng constant o 0.0 N/m. The speed o the bloc as t leaves the sprng s: Apply the conservaton o the mechancal energy: E 0 K K U U 0 G G ( mv 0) 0 x 0 0 v x (0.07).56 m/s. m () 6

22 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 Extra Problems Example: A sprng s compressed a dstance o cm by a 675 g ball. The sprng constant s 5 N/m. The ball s on a smooth surace as shown. The sprng s released and sets the ball nto moton. Fnd: (a) the speed o the ball when t leaves the sprng, (b) the speed o the ball just beore t leaves the edge o the table, (c) the netc energy o the ball just beore t hts the dec, (d) the horzontal dstance, x, t travels when t leaves the table untl t hts the dec. Assume that the ball rolls down the ramp..7 m.0 m m/s m (a) x mv v x (b) mv0 mgy 0 mv v v 0 gy0 (c) (d) mgy x x v m/s 0 0 K mgy0 x 0 g J y.0 y at t 0.45 s a 9.8 m x vt s.9 m s x 7

23 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 Example: A bloc, o mass m, s restng on a horzontal surace wth netc rcton coecent. The bloc s attached to an un-stretched sprng wth sprng constant (see Fgure). A orce F parallel to the surace s appled to the bloc. What s the speed o the bloc when ts dsplaced s d rom ts ntal poston? Method : use W U x sprng sprng K m v v W W W Fd x mg ( ) sprng F rcton Method : Use the equaton o moton ( ma F xdx mg F x mg use the nematc equaton v v a. d F x mg d W U W F sprng d 0 m m Example: A -g bloc s restng on a horzontal rctonless surace. The bloc s attached to an un-stretched sprng 800 N/m (see Fgure). A orce F = 80 N parallel to the surace s appled to the bloc. What s the speed o the bloc when t s dsplaced by 3 cm rom ts ntal poston? ), then Note that: W U x, or use the equaton o moton ( a F xdx )! sprng ( ) sprng mv K m v v sprng F W W Fd x (0.3) 3.64 J 3.64 v 0.78 m/s Example: A 0.50 g bloc attached to an deal sprng wth a sprng constant o 80 N/m oscllates on a horzontal rctonless surace. The speed o the bloc s 0.50 m/s, when the sprng s stretched by 4.0 cm. The maxmum speed the bloc can have s: mv mv x v m/s Example: A 4.0 g bloc s ntally movng to the rght on a horzontal rctonless surace at a speed o 5.0 m/s. It then compresses a horzontal sprng o sprng constant 00 N/m. At the nstant when the netc energy o the bloc s equal to the potental energy o the sprng, the mechancal energy o the bloc-sprng system s: d 0 8

24 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 E E E E mv Q: A 3.00 g bloc s dropped rom a heght o 40 cm onto a sprng o sprng constant (see Fgure). I the maxmum dstance the sprng s compressed = 0.30 m, nd. 0 (or at any place) J. E E E K K U U G G U U S S 0; mg 40 ( mg x ) 0 x N/m Example: As shown n Fgure, a 3.50 g bloc s accelerated rom rest by compressng aganst a sprng by a dstance x. The other end o the sprng s xed to the wall and the sprng has a sprng constant o 34 N/m. The bloc leaves the sprng and then travels over a rough horzontal loor wth a coecent o netc rcton The rctonal orce stops the bloc n dstance D = 7.80 m. Fnd the dstance x? Draw the ntal and nal states and nd the ollowng Intal state: (compressed sprng and statonary mass), K 0, U x, K 0, U 0 E x, S, S, m, m Fnal state: (un-stretched sprng and statonary mass at the end o the rough horzontal loor), K 0, U 0, K 0, U 0 E 0 Then use the equaton W. D mgd E x s, S, S, m, m.00 m. mgd x x Example8: A bloc (mass = 0.00 g) s pushed aganst a vertcal sprng compressng the sprng a dstance o h = 8.00 cm (see the gure a). The bloc s not attached to the sprng. When released rom rest, the bloc rses to a maxmum heght o H = m. Calculate the sprng constant. I the sprng s compressed a dstance h, we can apply the conservaton o mechancal energy; 9

25 Pro. Dr. I. Nasser Chapter 8_II_sprng November 3, 07 At the ntal state, At the nal state, K K K K 0 0, U U U mgh,mass,sprng,mass,sprng E mgh h,mass,sprng,mass,sprng h K K 0 0, U U U mg( H h) 0 E mg( H h) mg h E 0 E E H h mgh 0 mgh N/m h

26 Pro. Dr. I. Nasser Chapter 8_II_nosprng November 3, 07 Chapter 8 Revew Problems The ollowng three problems are related to Sec 8.4. The ncrease n thermal energy Eth d Wor-energy theorem agan! Start wth v v a. d, and use the Newton s second law to nd v v F F. d m K K K m v v W W Wth constant velocty, K 0 W W 0 a F, - Wor done, W F, by appled orce F s gven by the relaton W F. d F cos d F d where s the angle between F and the dsplacement d. = component o orce along the drecton o dsplacement, F p - The wor-energy theorem s wrtten as: Proo: Start wth F W F F. dstance. p K K K mv mv W p j j v v a. d, and use the Newton s second law a F, to nd v v F F. d K K K mv v W W m Constant velocty, K 0 W W 0 Note that: Accordng to the wor-energy theorem - wor done by all orces = ΔK - the wor s postve K Kand the velocty ncreases, v > v - the wor s negatve K K and the velocty decreases v < v v- only when external orce acts on an object ts energy ncreases by an amount equal to the wor done by ths orce. v- we can consder the netc energy o a body as the wor that the body do n comng to rest. Remember that: In the ollowng, eld could be reer to gravty, sprng, electrc, W W Feld ext U W W Feld Feld ext E E W non conservatve

27 Pro. Dr. I. Nasser Chapter 8_II_nosprng November 3, 07 Q: A 5.0 g bloc starts up a 30 o nclne wth 30 J o netc energy. How ar wll t slde up the nclne the coecent o netc rcton between the bloc and the nclne s 0.30? We wll suppose that the bloc slde up a dstance X along the nclne Q: A worer pushed a 7 g bloc 9. m along a level loor at constant speed wth a orce, F, drected wth angle 3 below the horzontal, see the gure. I the coecent o netc rcton between bloc and loor was 0.0, what were (a) the wor done by the worer s orce and (b) the ncrease n thermal energy o the bloc loor system? Snce the velocty s constant, are: y: F F sn mg, () a 0 and the x and the y component o Newton's second law N x: F cos () F F sn mg () N where m s the mass o the bloc, F s the orce exerted by the rope, and s the angle between that orce and the horzontal. Equatng () and (), one nds: mg F 7 N cos sn cos 3 0. sn 3 (a) The wor done on the bloc by the worer s, W Fdcos (7N)(9.m)cos J. (b) Snce F F sn mg = 60 N, we nd N E d th (60 N)(9.m) J Q: A rope s used to pull a 3.57 g bloc at constant speed 4.06 m along a horzontal loor. The orce on the bloc rom the rope s F 7.68 N and drected 5.0 above the horzontal. What are (a) the wor done by the rope's orce, (b) the ncrease n thermal energy o the bloc-loor system, and (c) the coecent o netc rcton between the bloc and loor? (a) The wor done on the bloc by the orce n the rope s, W Fd cos (7.68 N)(4.06m)cos J. (b) Usng or the magntude o the netc rcton orce, reveals that the ncrease n thermal energy s

28 Pro. Dr. I. Nasser Chapter 8_II_nosprng November 3, 07 Eth d (7.4 N)(4.06m) 30.J. (c) We can use Newton's second law o moton to obtan the rctonal and normal orces, then use / FN to obtan the coecent o rcton. Place the x axs along the path o the bloc and the y axs normal to the loor. The x and the y component o Newton's second law are x: F cos = 0 y: FN + F sn mg = 0, where m s the mass o the bloc, F s the orce exerted by the rope, and s the angle between that orce and the horzontal. The rst equaton gves = F cos = (7.68 N) cos5.0 = 7.4 N and the second gves FN mg F sn = (3.57 g)(9.8 m/s ) (7.68 N)sn5.0 = 33.0 N. Thus, 7.4 N 0.5. FN 33.0 N Example: A 40 g box ntally at rest s pushed 5.0 m along a rough horzontal loor wth a constant appled horzontal orce o 30 N. I the coecent o rcton between the box and loor s 0.30, nd (a) the wor done by the appled orce, (b) the energy lost due to rcton, (c) the change n netc energy o the box, and (d) the nal speed o the box. Fgure : (a) Appled orce and rcton orce both do wor on the box. (b) Dagram showng all the orces actng on the box. (a) The moton o the box and the orces whch do wor on t are shown n Fg. (a). The (constant) appled orce ponts n the same drecton as the dsplacement. Our ormula or the wor done by a constant orce gves The appled orce does J o wor. (b) Fg. (b) shows all the orces actng on the box. The vertcal orces actng on the box are gravty (mg, downward) and the loor s normal orce (N, upward). It ollows that N = mg and so the magntude o the rcton orce s The rcton orce s drected opposte the drecton o moton 80 and so the wor that t does s 3

29 Pro. Dr. I. Nasser Chapter 8_II_nosprng November 3, 07 or we mght say that J s lost to rcton, and The ncrease n thermal energy wll be E d J th (c) Snce the normal orce and gravty do no wor on the box as t moves, the net wor done s By the Wor Knetc Energy Theorem, ths s equal to the change n netc energy o the box: (d) Here, the ntal netc energy K was zero because the box was ntally at rest. So we have K = 6 J. From the denton o netc energy, K = ½ mv, we get the nal speed o the box: so that Example6: At t = 0 a.0 g ball s thrown rom a tall tower wth ˆ ˆ 0 U v 8 4 j m/s. What s o the ball Earth system between t = 0 and t = 6.0 s (stll ree all)? Snce tme does not drectly enter nto the energy ormulatons, we return to the nematc equatons to nd the change o heght durng ths t = 6.0 s lght. y v0 t gt y Ths leads to y 3 m. Thereore, U mgy 38 J 3. 0 J v 8 ˆ 0 ˆj m/s. What s Example7: A.00 g ball s thrown wth an ntal velocty o the maxmum change n the potental energy o the ball-earth system durng ts lght? K m( v v ) UG UG m( v ) (8 0 8 ) 00 J v Example: A small bloc o mass m begns rom rest at the top o a curved trac at a heght h and travels around a crcular loop o radus R. There s neglgble rcton between the bloc and the trac between ponts A and D, but the coecent o netc rcton on the horzontal surace between ponts D and E s μ. The dstance between ponts D and E s x. Answer all o the ollowng questons n terms o the gven quanttes and undamental constants. (a) Determne the speed o the bloc at pont B, at the bottom o the loop. (b) Determne the netc energy o the bloc at pont C, at the top o the loop. (c) Ater the bloc sldes down the loop rom pont C to pont D, t enters the rough porton o the trac. The speed o the bloc at pont E s hal the speed o the bloc at pont D. 4

30 Pro. Dr. I. Nasser Chapter 8_II_nosprng November 3, 07 Determne the speed o the bloc at pont D, just beore t enters the rough porton o the trac. (d) Determne the amount o wor done by rcton between ponts D and E. (e) Fnd an expresson or the coecent o netc rcton μ. Soluton Snce there s no rcton on the trac between ponts A and B, there s no loss o energy. Thus, (b) Conservaton o energy: U U K U A KB mgh mvb vb gh A C C mgh mg R K K mgh mgr C (c) There s no energy lost on the trac between ponts A and D, so the speed o the bloc at pont D s the same as the speed at pont B: v D gh (d) The wor done by rcton s the product o the rctonal orce and the dsplacement through whch t acts. W x F x mgx No use W m v D K v vd mgx N C gh h mgx mgx x (e) The rctonal orce causes the bloc to have a negatve acceleraton accordng to Newton s second law: mg a g m m Usng a nematc equaton, v v ax E D gh gh h 0 g x gx x Example: A worer does 500 J o wor n movng a 0 g box a dstance D on a rough horzontal loor. The box starts rom rest and ts nal velocty ater movng the dstance D s 4.0 m/s. Fnd the wor done by the rcton between the box and the loor n movng the dstance D. Total wor by the man s used to change the K.E. o the body and to overcome the rcton. So, K W W man W K Wman mv J Example: A.0 g bloc s released rom rest 60 m above the ground. Tae the gravtatonal potental energy o the bloc to be zero at the ground. At what heght above the ground s the netc energy o the bloc equal to hal ts gravtatonal potental energy? (Ignore ar resstance) 5

31 Pro. Dr. I. Nasser Chapter 8_II_nosprng November 3, 07 Apply the conservaton o the mechancal energy at the begnnng o the trp and at the hghest pont o the loop, we nd: K U K U 0 mgh mgx mgx x h 40 m Example: A. g bloc starts rom rest on a rough nclned plane that maes an angle o 30 above the horzontal. The coecent o netc rcton s 0.5. As the bloc moves 3.0 m down the plane, the change n the mechancal energy o the bloc s: E W mg cos30 d J Example: A 0.0 g bloc s released rom rest at 00 m above the ground. When t has allen 50 m, ts netc energy s: K U g mgh J Example: A 4.0 g bloc s ntally movng to the rght on a horzontal rctonless surace at a speed o 5.0 m/s. It then compresses a horzontal sprng o sprng constant 00 N/m. At the nstant when the netc energy o the bloc s equal to the potental energy o the sprng, the mechancal energy o the bloc-sprng system s: E 0 E E (or at any place) E mv J Example: A 5.0 g bloc starts up a 30 nclne wth 98 J o netc energy. The bloc sldes up the nclne and stops ater travelng 4.0 m. The wor done by the orce o rcton between the bloc and the nclne s: W E K U ( K K ) ( U U ) (0 98) ( sn 30 ) 00 J Example: Conservaton o Mechancal Energy Q: A 49 g projectle has a netc energy o 85 J when t s red at an angle o 3.0. Fnd (a) the tme o lght or the projectle and (b) the maxmum range o the projectle. We need to now the ntal velocty usng ts netc energy: g m K m K mv v 84 m 49 g s s 6

32 Pro. Dr. I. Nasser Chapter 8_II_nosprng November 3, 07 m o m vy0 vsn 84 sn s s m o m vx0 v cos 84 cos s s vy vy (a) Fnd the tme: vy vy0 gt t 4.7 s g 9.8 m (b) x vt 69 5 s = 500 m s Example: A 0.0 g object moves along a straght lne. The net orce actng on the object vares wth the object's dsplacement as shown n the graph. The object starts rom rest at dsplacement x = 0 and tme t = 0 and s dsplaced a dstance o 0 m. Determne each o the ollowng. a. The acceleraton o the partcle when ts dsplacement x = 6 m. The orce s constant rom tme zero tll ts dsplacement s 6 meters, so we can use the second law. F ma F g m m a 4 0 m s 0.0 g s b. The tme taen or the object to be dsplaced the rst m. Agan, the orce s constant rom the start tll the dsplacement s m. Ths means that the acceleraton s also constant, so we can use one o the nematc equatons to nd the tme. x ( m ) x xo vot at x at t.s a m 0 s c. The amount o wor done by the net orce n dsplacng the object the rst m. The wor done s the area under the curve: W(0 ) 4 N m 48 J d. The speed o the object at dsplacement x = m. We can use conservaton o energy to solve ths bt. g m m (48 ) W s m W (0 ) mv v.9 m 0.0g s e. The nal speed o the object at dsplacement x = 0 m. We can use conservaton o energy agan. The wor s equal to the area under the curve, so we add the area o the rectangle to the area o the trangle. W 4 N 0 m 4N m 64J g m m (64 ) W mv v s 5.3 m 0.0 g s W m 7

33 Pro. Dr. I. Nasser Chapter 8_II_nosprng November 3, 07 8

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