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

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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?

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Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?

• Energy balance with loss due to friction

at 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)v²

v² = 194/2.25 = 86.22

v = 9.29 m/s ANS-1

over rough surface friction force extracts work from box energy total

Wf = FF x 2.4 {where Wf = work done by friction force of FF)

FF = (0.27) x normal force

normal force = box weight = mg = (4.5)(9.8) = 44.1 N

FF = (0.27)(44.1) = 11.9 N

Wf = (11.9)(2.4) = 28.6 J

box speed just before hitting spring is given by K.E. of box

K.E. before spring = K.E.bottom – Wf = 194 – 28.6 = 165.4 J

K.E. before spring = 1/2 mv²

165.4 = (0.5)(4.5)v²

v² = 165.4/2.25 = 73.51

v = 8.57 m/s ANS-2

K.E. box = 165.4 J is stored at max spring compression as P.E. spring

165.4 = P.E. spring = 1/2 kx²

165.4 = (0.5)(500)x²

x² = 165.4/250 = 0.662

x = 0.81 m ANS-3

Each time box crosses rough surface it loses Wf = 28.6 J

Box starts with a total energy = 194 J

Box can cross rough surface = 194/28.6 = 6.8 times 