A 2.4 kg piece of wood slides on the surface shown in the figure (Figure 1) . Th
ID: 1537205 • Letter: A
Question
A 2.4 kg piece of wood slides on the surface shown in the figure (Figure 1) . The curved sides are perfectly smooth, but the rough horizontal bottom is 35 m long and has a kinetic friction coefficient of 0.20 with the wood. The piece of wood starts from rest 4.0 m above the rough bottom.
Part A
Where will this wood eventually come to rest?
Express your answer using two significant figures.
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Part B
For the motion from the initial release until the piece of wood comes to rest, what is the total amount of work done by friction?
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A 2.4 kg piece of wood slides on the surface shown in the figure (Figure 1) . The curved sides are perfectly smooth, but the rough horizontal bottom is 35 m long and has a kinetic friction coefficient of 0.20 with the wood. The piece of wood starts from rest 4.0 m above the rough bottom.
Part A
Where will this wood eventually come to rest?
Express your answer using two significant figures.
s = mSubmitMy AnswersGive Up
Part B
For the motion from the initial release until the piece of wood comes to rest, what is the total amount of work done by friction?
W = JSubmitMy AnswersGive Up
Explanation / Answer
Given
mass of piece m = 2.4 kg,
length of rough horizontal bottom surface is l = 35 m
kinetic friction coefficient of 0.20 with the wood,(mue_k)
piece of wood starts from rest from a height h = 4 m above the surface
Part A
here the energy equation is
k.e+p.e + Wother = 0
wood will comes to rest when w other = p.e
that is mue_k*mg*s = mgh
s = h/(mue_k)
s = 4/(0.2) = 20 m
so the wood will come to rest after displacing 20 m on the rough surface
PArt B
work done by the frictional force is
W other = mue_k*mg*s = 0.2*2.4*9.8*20 J = 94.08 J
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