Kinetic and heat energy problem

temaire
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Homework Statement


If a box is pushed along a level surface, what could be said if the coefficient of fricion([tex]\mu[/tex]) increases?

a) Final kinetic energy increases and heat energy increases
b) Final kinetic energy increases and heat energy decreases
c) Final kinetic energy decrease and heat energy increases
d) Final kinetic energy decreases and heat energy decreases

Homework Equations


[tex]E_k = \frac{mv_2} {2}[/tex]

[tex]F_F = \mu F_N[/tex]

[tex]W = Fd[/tex]

The Attempt at a Solution


I think that the answer is a), because if the coefficient of friction increases, the force of friction increases, which means that more work would be needed to overcome this friction. Therefore, kinetic energy will increase since it is equivalent to work. Also, heat energy increases because if the force of friction increases it is general knowledge that more heat will be produced. Am I right?
 
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It all depends how the box is pushed. If it's just given an initial speed and let go, then (c) is correct. Part of the KE is converted to heat energy.
 
The box has an initial speed of zero, and is pushed across the distance the entire way. I just forgot to mention that.
 
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How is it pushed? With constant speed? With constant accn? What is meant by "the co-eff of friction increases"? Does it increase along the path? Or are you comparing two separate cases when the co-effs are different?

Please formulate the problem precisely.
 
Well, this is a question I had on a test, so I can't remember the question word for word, but I think that the box is pushed with constant acceleration and that you are comparing two separate cases when the coefficients are different. My question wasn't very precise itself, so I might not be completely right.
 
If the actual accn in both cases are same, then obviously the box has same speed and hence same KE at the end in both cases. But the heat generated is more when the co-eff of friction is more.

A more practical question would be when they are pushed with the same force. Then the right ans would be (c) again, don't you think?
 
Actually I just remembered that the coeffecient of friction was increasing while the box was being pushed.
 
Well, then final KE decreases with respect to what? With respect to the case where the co-eff would have stayed constant, presumably. And we still don't know which is constant -- force, accn or speed or something else!

Anyway, since increase in the co-eff of friction always lead to generation of more heat at the expense of the loss of KE, the ans (c) still is the most reasonable one.

Enough about this one.
 

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