# Heat due to friction problem

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## Homework Statement

A metal block of mass 10 kg is dragged on a rough horizontal surface with a constant speed of 5 m/s. If the coefficient of kinetic friction between the block and the road is 0.4, then the rate of generation of heat is. take g=10 m/s^2?

## Homework Equations

F=ma
Frictional force=μN

## The Attempt at a Solution

So I thought, the heat generated is the lost kinetic energy per second (if we stop applying the force), so I calculated the time it takes for the block to reach 0 velocity, the retardation is (0.4 x 100)/10 from v=u+at I got time=5/4 seconds. the rate should be total KE/time but this is incorrect, can someone give an intuitive explanation on how to solve this? Thank you

## Answers and Replies

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kuruman
Homework Helper
Gold Member
Rate of heat generation is really power. Can you relate power and force and velocity?

Gold Member
Rate of heat generation is really power. Can you relate power and force and velocity?
Power is Force x velocity. Is this the energy we're constantly providing to keep it moving at 5 m/s?

kuruman
Homework Helper
Gold Member
If you use the pulling force in the expression, it is the power generated by that force. If you use the force of friction, it is the power lost to friction.

Gold Member
If you use the pulling force in the expression, it is the power generated by that force. If you use the force of friction, it is the power lost to friction.
Ok, I understand this. Why is my approach wrong though?

kuruman
It is wrong because when you say that P = KE/t, you are assuming that in equal times you are losing equal amounts of energy. That is, you are assuming that from t = 0 to t = 1 s you are losing the same amount of KE as from t = 1 s to t = 2 s. This is not the case when the block loses kinetic energy until it stops. The correct expression for the power loss is $P=\frac{d(KE)}{dt}=\frac{d}{dt}\left( \frac{1}{2}mv^2 \right)=m v \frac{dv}{dt}=mav = F_{net}v.$
It is wrong because when you say that P = KE/t, you are assuming that in equal times you are losing equal amounts of energy. That is, you are assuming that from t = 0 to t = 1 s you are losing the same amount of KE as from t = 1 s to t = 2 s. This is not the case when the block loses kinetic energy until it stops. The correct expression for the power loss is $P=\frac{d(KE)}{dt}=\frac{d}{dt}\left( \frac{1}{2}mv^2 \right)=m v \frac{dv}{dt}=mav = F_{net}v.$