Question

A 100 kg block is started with a speed of $2.0m{s}^{-1}$ on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20.

Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt

• A. 100 J
• B. 0 J
• C. 200 J
• D. 400 J

Answer 1

The correct option is C

200 J

The internal energy of a system is the total kinetic energy available in the system.
Initially when the block is moving with a velocity of 2 m/s, it poses a kinetic energy of
$\left(\frac{1}{2}\right)m{v}^2$, wherem = mass of block
v = velocity of block

$=\left(\frac{1}{2}\right)\times 100\times 2^2=200J$.

The belt which is stationary doesn't possess any kinetic energy.
Hence the total internal energy of the belt block system initially is 200 J.
Finally when the block comes to rest, which it will due to friction, the total kinetic energy of the system reduces to zero.
Thus the total change in internal energy of the belt-block system is 200 J.

This 200 J is lost as heat given out by the system. Some amount of heat will be used to increase the temperature of the belt and the block, but this heat will also be eventually lost to the atmosphere, which will be at a lower temperature.