A 100 kg block is given an initial speed of 2.0 m s^ -1 on a long,...

The initial kinetic energy of the block is given by $K_i = \dfrac{1}{2}mv^2$.

Substituting the given values, $K_i = \dfrac{1}{2} \times 100 \times (2.0)^2 = 200 \text{ J}$.

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 $\Delta U = -\Delta K = -(K_f - K_i)$.

$\Delta U = -(0 - 200) = 200 \text{ J}$.