kean (msk955) – Homework 04 – yao – (56725) 1This print-out should have 26 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsA force F is exerted by a broom handle on thehead of the broom, which has a mass m. Thehandle is at an angle θ to the horizontal, asshown bel ow.FθWhat is the work done by the force on thehead of the broom as it moves a dist ance dacross a hori zontal floor?1. W = F m d sin θ2. W = F d sin θ3. W = F m tan θ4. W = F d cos θ correct5. W = F m cos θExplanation:By definition, the work done by a force isgiven by the formulaW =~F ·~d = F d cos θ,where~d is the displacement vector, and θ isthe angle between~F and~d.Simply apply the definition, the work isW = F d cos θ .002 10.0 pointsA 6 .44 g bullet moving at 657 m/s penetratesa t r ee to a depth of 3.01 cm.Use energy considerations to find the aver-age frictional force that stops the bullet.Correct answer: 46176.4 N .Explanation:Applying the work-energy theorem,Wf= ∆K =~f · ~s = −f s0 −12m v2= −f sf =m v22 s=(0.00644 kg)(657 m/s)22(0.0301 m)= 46176 .4 N .003 10.0 pointsA block sliding on a horizontal surface has aninitial speed of 0.5 m/s. The block travels adistance of 1 m as it slows to a stop.What distance would the block have trav-eled if its initial speed had been 1 m/s?1. 3 m2. 2 m3. 4 m correct4. more information is needed to answer thequestion5. 1 m6. 0.5 mExplanation:Wnc= −fkL = 0 −12m v02−µ m g L = −12m v02L =v022 µ gSo the distance is propo r tional to the squareof the initial speed.004 10.0 pointsKaren has a mass of 49.5 kg as she ridesthe up escalator at Woodley Park Station ofthe Washington D.C. Metro. Karen rode adistance of 69.1 m, the longest escalator inthe free world.kean (msk955) – Homework 04 – yao – (56725) 2The acceleration of gravity is 9.8 m/s2.How much work did the escalator do onKaren if it has an inclination of 29.1◦?Correct answer: 16302.2 J.Explanation:dhθThe escalator raised Karen a height ofh = d sin θso the work done on Karen by the escalator isW = mgh= mgd sin θ= (49.5 kg) (9.8 m/s2) (69.1 m) sin 29.1◦= 16302.2 J005 10.0 pointsFive ramps lead from the ground to the secondfloor of a workshop, as sketched below. Allfive ramps have the same height; ramps B,C, D and E have the same length; ramp A islonger than the other four. You need to pusha heavy cart up to the second floor and youmay choose any one of the five ramps.ABCDEAssuming no frictional forces on the cart,which ramp would require you t o do the leastwo rk?1. Unable to determine without knowing theexact profiles of ramps C, D or E.2. Same work for t he straight ramps A andB; less work for ramps C, D, and E.3. Ramp D.4. Ramp B.5. Same work for all five ramps. correct6. Ramp A.7. Ramp C.8. Ramp E.9. Same work for ramps B, C, D or E; morewo rk for r amp A.Explanation:Let h be the height of the ramps. Sincethere is no friction, we can use the work-energy theorem.Wtot= Wperson+ Wgravity= ∆KHence for all the ramps,Wperson= −Wgravity+ ∆K= mgh + ∆KIn particular, if ∆K = 0 (the cart sta rts fromrest and ends at rest), Wperson= mgh, for allthe ramps.006 10.0 pointsA force F of magnit ude 2x3is applied tostop a particle moving with an initial velocityv0. The parti cle travels from x = 0 to x = Din coming to a stop after the force F is applied.The work done by F is1. −D32. −12D4correct3.12D34. D45. −12D3kean (msk955) – Homework 04 – yao – (56725) 36. 2D37. −D48.12D4Explanation:Because the force is to stop the particle, it ’sopposite to the particle’s motion direction,i.e., the force is negative. From the definitionof work,W =ZF ds =ZD0−2x3dx = −12D4007 10.0 pointsA woman pushes a lawn mower with a con-stant force of 127 N a t an angle of 33◦withrespect to the horizontal. The lawn mowermoves at a speed of 1 1 cm/sec.What is her power output? (1 hp = 746Watts)Correct answer: 0.0157054 hp.Explanation:P = F (cos θ) v= (127 N) (cos 33◦) (0.11 m/sec)= 11.7162 W= (11.7162 W)1 hp746 W= 0.0157054 hp008 10.0 pointsWhen a rattlesnake strikes, its head accel-erates from rest to a speed of 27 m/s in0.67 seconds. Assume for si mpl icity that theonly moving part of the snake is its head ofmass 150 g.How much (average) power does the rat-tlesnake need to accelerate its head that fast?Correct answer: 81.6045 W.Explanation:As the snake’ s head accelerates, its kineticenergy increases from zero toE =12m v2=12(150 g) (27 m/s)2= 54.675 J .To deliver that much energy in t = 0.67 s, thesnake needs powerP =Et=54.675 J0.67 s=81.6045 W .009 10.0 pointsA 125 kg physics professor has fallen intothe Grand Canyon. Luckily, he managedto g rab a branch and is now hanging 51 mbelow the rim. A student (majoring in lin-guistics and physi cs) decides to perform arescue/experi ment using a nearby horse. Af-ter lowering a rope to her fallen hero andattaching the other end to the horse, the stu-dent measures how long it takes for the horseto pull the fallen physicist to the rim of theGrand Canyon.The acceleration of gravity is 9.8 m/s2.If the horse’s output power is truly 1 horse-power (746 W), and no energy is lost t o fric-tion, how long should the process take?Correct answer: 83.7466 s.Explanation:The work against gr avity which is necessaryto raise the professor to the rim of the canyonis given byW = m g ∆h= (125 kg) (9.8 m/s2) (51 m)= 62475 J .If the horse’s power output is one horsepower,then the the time for the horse to provide thenecessary work is given bytup=WPoutput=62475 J746 W= 83.7466 s .010 10.0 pointskean (msk955) – Homework 04 – yao – (56725) 4When an automobile moves with constant ve-locity, the power developed is used to over-come the frictional forces exerted by the airand the road.If the engine develops 8 hp, what tota l fric-tional force acts on t he car at 72 mph? Onehorsep ower equals 746 W, and o ne mile is1609 m.Correct answer: 185.457 N .Explanation:ApplyP =Et=F dt= F vF =Pv=8 hp ·746 Whp72 mph ·1609 mmi·h3600 s= 185.457 N011 10.0 pointsWater flows over a section of Niagar a Falls ata rate of 1.18 × 106kg/s a nd falls 25.2 m.The acceleration of gravity is 9.8 m/s2.How much power is generated by the fallingwa ter?Correct answer: 291.413 MW.Explanation:Given : r = 1.18 × 106kg/s andh = 25.2 m .The power of the water is equal to thechange in potent ial energy per unit ti m e, soP =∆U∆t=∆m g h∆t=∆m∆tg h= (1.18 × 106kg/s) (9.8 m/s2) (25.2 m)×1 MW106W=