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Dana Herzog

Feb 20, 2025

How do you do this work and energy problem? physics urgent help!?

A 4.5 kg box slides down a 4.4-m-high frictionless hill, starting from rest, across a 2.4m-wide horizontal surface, then hits a horizontal spring with spring constant 500N/m . The other end of the spring is anchored against a wall. The ground under the spring is frictionless, but the 2.4-m-long horizontal surface is rough. The coefficient of kinetic friction of the box on this surface is 0.27.What is the speed of the box just before reaching the rough surface?What is the speed of the box just before hitting the spring?How far is the spring compressed?Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?

1 Answers

Energy balance with loss due to frictionat top of frictionless hill the box has its TOTAL starting energy as P.E.P.E. top = mgh {where m = 4.5, g = 9.8, h = 4.4)P.E. top = (4.5)(9.8)(4.4) = 194 J {total energy at start}at bottom of frictionless hill the box has its TOTAL energy as K.E.P.E. top = K.E. bottom = 1/2 mv² 194 = 0.5(4.5)v²v² = 194/2.25 = 86.22v = 9.29 m/s ANS-1over rough surface friction force extracts work from box energy totalWf = FF x 2.4 {where Wf = work done by friction force of FF)FF = (0.27) x normal forcenormal force = box weight = mg = (4.5)(9.8) = 44.1 NFF = (0.27)(44.1) = 11.9 NWf = (11.9)(2.4) = 28.6 Jbox speed just before hitting spring is given by K.E. of boxK.E. before spring = K.E.bottom - Wf = 194 - 28.6 = 165.4 JK.E. before spring = 1/2 mv² 165.4 = (0.5)(4.5)v² v² = 165.4/2.25 = 73.51v = 8.57 m/s ANS-2K.E. box = 165.4 J is stored at max spring compression as P.E. spring165.4 = P.E. spring = 1/2 kx²165.4 = (0.5)(500)x²x² = 165.4/250 = 0.662x = 0.81 m ANS-3Each time box crosses rough surface it loses Wf = 28.6 JBox starts with a total energy = 194 JBox can cross rough surface = 194/28.6 = 6.8 timessince question asks for "complete trips" answer is 6 times ANS-4... Show More

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