Physics 4A Solutions to Chapter 7/8 Homework

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1 Chapter 7 Questons: 6, 8 Exercses & Problems 15, 17, 3, 37, 49, 6, 76 Physcs 4A Solutons to Chapter 7/8 Homework Chapter 8 Questons:, 4 Exercses & Problems: 14, 1, 3, 4, 7, 9, 57, 6, 7 Answers to Questons: Q 7-6 (a) 3 m (b) 3 m (c) and 6 m (d) -x Q 7-8 (a) A, F ; B, F 1; C, F 3; D, F 4 (b) E, A and D; F, B and C; G and H meanngless because K cannot have negatve values Q 8- (a) AB, CD, then BC and DE te (zero force) (b) 5 J (c) 5 J (d) 6 J (e) FG (f) DE Q 8-4 (a) 4 (b) returns to ts startng pont and repeats the trp (c) 1 (d) 1 Answers to Problems: P 7-15 (a) The forces are constant, so the work done by any one of them s gven by = F d, where d s the dsplacement. Force F 1 s n the drecton of the dsplacement, so = Fdcos φ = (5. N)(3. m)cos = 15. J. Force F makes an angle of 1 wth the dsplacement, so Force F 3 = F dcos φ = (9. N)(3. m)cos1 = 13.5 J. s perpendcular to the dsplacement, so The net work done by the three forces s 3 = F 3 d cos φ 3 = snce cos 9 =.

2 = = 15. J 13.5 J + =+ 1.5 J. (b) If no other forces do work on the box, ts knetc energy ncreases by 1.5 J durng the dsplacement. P 7-17 e use F to denote the upward force exerted by the cable on the astronaut. The force of the cable s upward and the force of gravty s mg downward. Furthermore, the acceleraton of the astronaut s a = g/1 upward. Accordng to Newton s second law, the force s gven by 11 F mg = ma F = m( g+ a) = mg, 1 n the same drecton as the dsplacement. On the other hand, the force of gravty has magntude F = mg and s opposte n drecton to the dsplacement. g (a) Snce the force of the cable F and the dsplacement done by F s d are n the same drecton, the work F 11mgd 11 (7 kg)(9.8 m/s )(15 m) 4 4 = Fd = = = J 1. 1 J. 1 1 (b) Usng Eq. 7-7, the work done by gravty s = F d = mgd = = 4 g g 4 (7 kg)(9.8 m/s )(15 m) J J (c) The total work done s the sum of the two works: = + = = net F g J J J J Snce the astronaut started from rest, the work-knetc energy theorem tells us that ths s her fnal knetc energy. 3 K (1.6 1 J) (d) Snce K = 1 mv, her fnal speed s v = = = 5.4 m/s. m 7 kg Note: For a general upward acceleraton a, the net work done s net = F + g = Fd Fgd = m( g+ a) d mgd = mad.

3 Snce =Δ K = mv net /, by the work-knetc energy theorem, the speed of the astronaut would be v= ad, whch s ndependent of the mass of the astronaut. P The work done by the sprng force s gven by Eq. 7-5: s = k ( x xf). Snce Fx = kx, the slope n Fg corresponds to the sprng constant k. Its value s gven by 3 k = 8 N/cm=8. 1 N/m. (a) hen the block moves from x =+ 8. cm to x = + 5. cm, we have 1 ( N/m)[(.8 m) (.5 m) ] 15.6 J 16 J. s = = (b) Movng from x =+ 8. cm to x = 5. cm, we have 1 ( N/m)[(.8 m) (.5 m) ] 15.6 J 16 J. s = = (c) Movng from x =+ 8. cm to x = 8. cm, we have 1 ( N/m)[(.8 m) (.8 m) ] J. s = = (d) Movng from x =+ 8. cm to x = 1. cm, we have 1 ( N/m)[(.8 m) (.1 m) ] 14.4 J 14 J. s = = P 7-37 (a) e frst multply the vertcal axs by the mass, so that t becomes a graph of the appled force. Now, addng the trangular and rectangular areas n the graph (for x 4) gves 4 J for the work done. (b) Countng the areas under the axs as negatve contrbutons, we fnd (for x 7) the work to be 3 J at x = 7. m. (c) And at x = 9. m, the work s 1 J. (d) Equaton 7-1 (along wth Eq. 7-1) leads to speed v = 6.5 m/s at x = 4. m. Returnng to the orgnal graph (where a was plotted) we note that (snce t started from rest) t has receved

4 acceleraton(s) (up to ths pont) only n the +x drecton and consequently must have a velocty vector pontng n the +x drecton at x = 4. m. (e) Now, usng the result of part (b) and Eq. 7-1 (along wth Eq. 7-1) we fnd the speed s 5.5 m/s at x = 7. m. Although t has experenced some deceleraton durng the x 7 nterval, ts velocty vector stll ponts n the +x drecton. (f) Fnally, usng the result of part (c) and Eq. 7-1 (along wth Eq. 7-1) we fnd ts speed v = 3.5 m/s at x = 9. m. It certanly has experenced a sgnfcant amount of deceleraton durng the x 9 nterval; nonetheless, ts velocty vector stll ponts n the +x drecton. P 7-49 e have a loaded elevator movng upward at a constant speed. The forces nvolved are: gravtatonal force on the elevator, gravtatonal force on the counterweght, and the force by the motor va cable. The total work s the sum of the work done by gravty on the elevator, the work done by gravty on the counterweght, and the work done by the motor on the system: = e + c + m. Snce the elevator moves at constant velocty, ts knetc energy does not change and accordng to the work-knetc energy theorem the total work done s zero, that s, = Δ K =. The elevator moves upward through 54 m, so the work done by gravty on t s e e 5 (1 kg)(9.8 m/s )(54 m) J. = m gd = = The counterweght moves downward the same dstance, so the work done by gravty on t s 5 = m gd = (95 kg)(9.8 m/s )(54 m) = J. c Snce =, the work done by the motor on the system s c m e c J J J. = = = 5 Ths work s done n a tme nterval of Δ t = 3. mn = 18 s, so the power suppled by the motor to lft the elevator s m J P = = = Δt 18 s P 7-6 (a) The compresson of the sprng s d =.1 m. The work done by the force of gravty (actng on the block) s, by Eq. 7-1,

5 c h 1 = mgd = (. 5kg) 9.8 m / s (.1 m) = 9. J. (b) The work done by the sprng s, by Eq. 7-6, 1 1 = kd = ( 5 N / m) (.1 m) = 18. J. (c) The speed v of the block just before t hts the sprng s found from the work-knetc energy theorem (Eq. 7-15): whch yelds v ΔK = 1 mv = + 1 ( )( + ) ( )(.9 J 1.8 J) m.5 kg 1 = = = 3.5 m/s. (d) If we nstead had v = 7 m/s, we reverse the above steps and solve for d. Recallng the theorem used n part (c), we have 1 1 mv = + = mgd kd 1 whch (choosng the postve root) leads to mg + m g + mkv d = k whch yelds d =.3 m. In order to obtan ths result, we have used more dgts n our ntermedate results than are shown above (so v = 1.48 m/s = m/s and v = 6.94 m/s). P 7-76 (a) The component of the force of gravty exerted on the ce block (of mass m) along the nclne s mg sn θ, where θ = sn b g gves the angle of nclnaton for the nclned plane. Snce the ce block sldes down wth unform velocty, the worker must exert a force F uphll wth a magntude equal to mg sn θ. Consequently,.91m F = mg θ = = 1.5m sn (45 kg)(9.8 m/s ).7 1 N. (b) Snce the downhll dsplacement s opposte to F, the work done by the worker s

6 c h 1 =. 7 1 = 4. 1 N (1.5 m) J. (c) Snce the dsplacement has a vertcally downward component of magntude.91 m (n the same drecton as the force of gravty), we fnd the work done by gravty to be c h = ( 45 kg) 9. 8 m / s (.91 m) = 4. 1 J. (d) Snce F N s perpendcular to the drecton of moton of the block, and cos9 =, work done by the normal force s 3 = by Eq (e) The resultant force F net s zero snce there s no acceleraton. Thus, ts work s zero, as can be checked by addng the above results + + =. 1 3 P 8-14 e use Eq. 8-18, representng the conservaton of mechancal energy (whch neglects frcton and other dsspatve effects). (a) The change n potental energy s ΔU = mgl as t goes to the hghest pont. Thus, we have ΔK+ ΔU = K K + mgl= whch, upon requrng K top =, gves K = mgl and thus leads to top v K m = = gl = (9.8 m/s )(.45 m) =.98 m/s. (b) e also found n Problem 9-4 that the potental energy change s ΔU = mgl n gong from the ntal pont to the lowest pont (the bottom). Thus, ΔK + ΔU = K K mgl= bottom whch, wth K = mgl, leads to K bottom = mgl. Therefore, v bottom K m bottom = = 4gL = 4(9.8 m/s )(.45 m) = 4.1 m/s. (c) Snce there s no change n heght (gong from ntal pont to the rghtmost pont), then ΔU =, whch mples ΔK =. Consequently, the speed s the same as what t was ntally, vrght = v =.98 m/s.

7 (d) It s evdent from the above manpulatons that the results do not depend on mass. Thus, a dfferent mass for the ball must lead to the same results. P 8-1 e use Eq. 8-18, representng the conservaton of mechancal energy (whch neglects frcton and other dsspatve effects). The reference poston for computng U (and heght h) s the lowest pont of the swng; t s also regarded as the fnal poston n our calculatons. (a) Careful examnaton of the fgure leads to the trgonometrc relaton h = L L cos θ when the angle s measured from vertcal as shown. Thus, the gravtatonal potental energy s U = mgl(1 cos θ ) at the poston shown n Fg. 8-3 (the ntal poston). Thus, we have K + U = Kf + U f 1 1 mv + mgl b1 cosθ g = mv + whch leads to 1 v= mv mgl(1 cos θ) v gl(1 cos θ) m + = + (8. m/s) (9.8 m/s )(1.5 m)(1 cos 4 ) 8.35 m/s. = + = (b) e look for the ntal speed requred to barely reach the horzontal poston descrbed by v h = and θ = 9 (or θ = 9, f one prefers, but snce cos( φ) = cos φ, the sgn of the angle s not a concern). whch yelds v gl θ b K + U = K + U 1 mv + mgl 1 cosθ = + mgl = cos = (9.8 m/s )(1.5 m)cos 4 = 4.33 m/s. g h h (c) For the cord to reman straght, then the centrpetal force (at the top) must be (at least) equal to gravtatonal force: mvt = mg mvt = mgl r where we recognze that r = L. e plug ths nto the expresson for the knetc energy (at the top, where θ = 18 ). K + U = K + U whch leads to b g b 1 1 mv + mgl 1 cosθ = mvt + mg 1 cos mv + mgl b1 cos θ g = ( mgl) + mg( L) t t g

8 v gl θ = (3+ cos ) = (9.8 m/s )(1.5 m)(3 + cos 4 ) = 7.45 m/s. (d) The more ntal potental energy there s, the less ntal knetc energy there needs to be, n order to reach the postons descrbed n parts (b) and (c). Increasng θ amounts to ncreasng U, so we see that a greater value of θ leads to smaller results for v n parts (b) and (c). P 8-3 (a) As the strng reaches ts lowest pont, ts orgnal potental energy U = mgl (measured relatve to the lowest pont) s converted nto knetc energy. Thus, 1 mgl = mv v = gl. th L = 1. m we obtan v = gl = (9.8 m/s )(1. m) = 4.85 m/s. (b) In ths case, the total mechancal energy s shared between knetc 1 mv b and potental mgy b. e note that y b = r where r = L d =.45 m. Energy conservaton leads to 1 mgl = mvb + mgyb b g. whch yelds vb = gl g r =.4 m s P 8-4 e denote m as the mass of the block, h =.4 m as the heght from whch t dropped (measured from the relaxed poston of the sprng), and x as the compresson of the sprng (measured downward so that t yelds a postve value). Our reference pont for the gravtatonal potental energy s the ntal poston of the block. The block drops a total dstance h + x, and the fnal gravtatonal potental energy s mg(h + x). The sprng potental energy s 1 kx n the fnal stuaton, and the knetc energy s zero both at the begnnng and end. Snce energy s conserved K + U = K + U f f = mg h + x + 1 ( ) kx whch s a second degree equaton n x. Usng the quadratc formula, ts soluton s b g mg ± mg + mghk x = k. Now mg = 19.6 N, h =.4 m, and k = 196 N m, and we choose the postve root so that x >.

9 x = 196 b gb gb g =.1 m. P 8-7 (a) To fnd out whether or not the vne breaks, t s suffcent to examne t at the moment Tarzan swngs through the lowest pont, whch s when the vne f t ddn't break would have the greatest tenson. Choosng upward postve, Newton's second law leads to T mg = m v r where r = 18. m and m= g = = 7. kg. e fnd the v from energy conservaton (where the reference poston for the potental energy s at the lowest pont). 1 mgh = mv v = gh where h = 3. m. Combnng these results, we have F HG T mg m gh h = + = mg 1+ r r whch yelds 933 N. Thus, the vne does not break. (b) Roundng to an approprate number of sgnfcant fgures, we see the maxmum tenson s roughly N. I K J P 8-9 e refer to ts startng pont as A, the pont where t frst comes nto contact wth the sprng as B, and the pont where the sprng s compressed x =.55 m as C, as shown n the fgure below. Pont C s our reference pont for computng gravtatonal potental energy. Elastc potental energy (of the sprng) s zero when the sprng s relaxed.

10 Informaton gven n the second sentence allows us to compute the sprng constant. From Hooke's law, we fnd F 7 N 4 k = = = N m. x. m The dstance between ponts A and B s l and we note that the total sldng dstance l + x s related to the ntal heght h A of the block (measured relatve to C) by where the nclne angle θ s 3. snθ = l ha + x (a) Mechancal energy conservaton leads to whch yelds h A 1 KA + UA = KC + UC + mgha = kx 4 kx ( N/m)(.55 m) = = =.174 m. mg (1 kg)(9.8 m/s ) Therefore, the total dstance traveled by the block before comng to a stop s l h.174 m + x = A = =.347 m.35 m. sn3 sn3 (b) From ths result, we fnd l = x =.347 m.55 m =.9 m, whch means that the block has descended a vertcal dstance Δ y = h h = l sn θ = (.9 m)sn3 =.146 m A B n sldng from pont A to pont B. Thus, usng Eq. 8-18, we have mgh = mv + mgh mv = mg Δ y A B B B whch yelds B = Δ = (9.8 m/s )(.146 m) = 1.69 m/s 1.7 m/s. v g y Note: Energy s conserved n the process. The total energy of the block at poston B s 1 1 (1 kg)(1.69 m/s) (1 kg)(9.8 m/s B B B )(.8 m).4 J E = mv + mgh = + =,

11 whch s equal to the elastc potental energy n the sprng: kx = ( N/m)(.55 m) =.4 J. P 8-57 Snce the valley s frctonless, the only reason for the speed beng less when t reaches the hgher level s the gan n potental energy ΔU = mgh where h = 1.1 m. Sldng along the rough surface of the hgher level, the block fnally stops snce ts remanng knetc energy has turned to thermal energy ΔEth = fkd = μ mgd, where μ =.6. Thus, Eq (wth = ) provdes us wth an equaton to solve for the dstance d: K = ΔU + ΔE th = mg h+ μd b g where K = mv / and v = 6. m/s. Dvdng by mass and rearrangng, we obtan d v h = = 1. m. μg μ P 8-6 e look for the dstance along the nclne d, whch s related to the heght ascended by Δh = d sn θ. By a force analyss of the style done n Chapter 6, we fnd the normal force has magntude F N = mg cosθ, whch means f k = μ k mg cosθ. Thus, Eq (wth = ) leads to = Kf K + ΔU + ΔE th = K + mgdsnθ + μkmgdcosθ whch leads to d K 18 = = mg sn + cos.. sn +. cos = 43m.. θ μ θ b k g b gb gb g P 8-7 (a) e take the gravtatonal potental energy of the sker-earth system to be zero when the sker s at the bottom of the peaks. The ntal potental energy s U = mgh, where m s the mass of the sker, and H s the heght of the hgher peak. The fnal potental energy s U f = mgh, where h s the heght of the lower peak. The sker ntally has a knetc energy of K =, and the fnal knetc energy s Kf = 1 mv, where v s the speed of the sker at the top of the lower peak. The normal force of the slope on the sker does no work and frcton s neglgble, so mechancal energy s conserved:

12 Thus, 1 U K U K mgh mgh mv + = f + f = +. = ( ) = (9.8 m/s )(85 m 75 m) = 44 m/s. v g H h (b) e recall from analyzng objects sldng down nclned planes that the normal force of the slope on the sker s gven by F N = mg cos θ, where θ s the angle of the slope from the horzontal, 3 for each of the slopes shown. The magntude of the force of frcton s gven by f = μ k F N = μ k mg cos θ. The thermal energy generated by the force of frcton s fd = μ k mgd cos θ, where d s the total dstance along the path. Snce the sker gets to the top of the lower peak wth no knetc energy, the ncrease n thermal energy s equal to the decrease n potental energy. That s, μ k mgd cos θ = mg(h h). Consequently, H h (85 m 75 m) μk = = =.36. d θ 3 cos (3. 1 m) cos 3

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