Work and Energy Problem

  • Thread starter smbascug
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  • #1
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A 60.0 kg crate, starting from rest, is pulled across the floor with a constant force of 100.0 N. For the first 10.0 m the floor is frictionless and for the next 10.0 m the coefficient of friction is 0.20. What is the final speed of the crate?

Is it possible to use KE + PE = KE + PE
1/2mv^2 + mgh = 1/2mv^2 + mgh
 

Answers and Replies

  • #2
Doc Al
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No, since energy is not conserved. Instead, consider the work done by the applied forces. (The work-energy theorem.)
 
  • #3
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How can you tell if energy is coserved or not?
 
  • #4
smbascug said:
How can you tell if energy is coserved or not?

Work is done, both by the puller and the frictional force. The friction generates heat, which certainly tells us that energy is not conserved.
 
  • #5
Doc Al
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smbascug said:
How can you tell if energy is coserved or not?
Mechanical energy is conserved if the only forces acting are conservative forces, like gravity. As Captain Zapp0 points out, here we have additional forces acting: the applied force and the friction. These forces do work on the object, changing its kinetic energy.

If you attempted to use the conservation equation you quoted, you'll quickly find that it gets you nowhere since the gravitational PE doesn't change. (We assume that the crate is moving across a horizontal floor.)
 
  • #6
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m=60.0kg u=0.20
F= 100.0 N
d1==10.0m

KE(final) - KE(initial)
1/2mv^2 - 1/2mv^2

umm... how do i isolate v? or is this even correct?
 
  • #7
Doc Al
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Not clear what you've done.

Find the work done by the forces during the first 10m and the second 10m. The total work done will equal the change in the crate's KE.
 

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