A 4.5 kg box slides down a 4.2-m -high frictionless hill, starting from rest, ac

Question

A 4.5 kg box slides down a 4.2-m -high frictionless hill, starting from rest, across a 2.0-m -wide horizontal surface, then hits a horizontal spring with spring constant 500 N/m . The other end of the spring is anchored against a wall. The ground under the spring is frictionless, but the 2.0-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? 9.1 m/s

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?

Explanation / Answer

At the top of the frictionless hill the box has its TOTAL starting energy which will be in the form of P.E.
P.E. top = mgh {where m = 4.5, g = 9.8, h = 4.2)
P.E. top = (4.5)(9.8)(4.2) = 185 J This is the total energy at the starting point
at bottom of frictionless hill the box has its TOTAL energy as K.E.
P.E. top = K.E. bottom = 1/2 mv2
185 = 0.5(4.5)v2
v2 = 185/2.25 = 82.32
v = 9.07 m/s
b) The work done by frictional force is
Wf = FF x 2
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) = 23.8 J
box speed just before hitting spring is given by K.E. of box
K.E. before spring = K.E.bottom - Wf = 185 - 23.8 = 161.2 J
K.E. before spring = 1/2 mv2
161.2 = (0.5)(4.5)v2
v2 = 161.2/2.25 = 71.644
v = 8.46c m/s
K.E. box = 161.2 J is stored at max spring compression as P.E. spring
161.2 = P.E. spring = 1/2 kx2
161.2 = (0.5)(500)x2
x2 = 161.2/250 = 0.806
x = 0.897 m
d) Each time box crosses rough surface it loses Wf = 23.8 J
Box starts with a total energy = 185 J
Box can cross rough surface = 185/23.8 = 7.77 times
since it asked for complete trips it will be 7 times because 7.77 means 8th time is not completed.

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