A 100 kg block is given an initial speed of 2.0 m 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 . Determine the change in the internal energy of the block–belt system as the block comes to a stop on the belt.

Step-by-step solution

The initial kinetic energy of the block is given by K_i = 1 2 mv^2 . Substituting the given values, K_i = 1 2 100 (2.0)^2 = 200 J . Since the block comes to a stop, its final kinetic energy is K_f = 0 . According to the law of conservation of energy for the isolated block-belt system, the loss in macroscopic mechanical energy is converted into the internal energy (thermal energy) of the system. Therefore, the change in internal energy is U = - K = -(K_f - K_i) . U = -(0 - 200) = 200 J .