A 100 kg block is started with a speed of 2.0 m s^ -1 on a long,...

Let the ground frame be $S$ and the moving frame be $S'$.

In the ground frame $S$, the initial velocity of the block is $v_{i} = 2.0 \text{ m s}^{-1}$ and the belt is stationary ($v_{\text{belt}} = 0$).

The frame $S'$ is moving with a velocity $v_{F} = 2.0 \text{ m s}^{-1}$ parallel to the block's initial velocity.

Initial velocity of the block in frame $S'$ is:
$u' = v_{i} - v_{F} = 2.0 - 2.0 = 0 \text{ m s}^{-1}$

When the block ceases slipping, it moves together with the stationary belt. The final velocity of the block in the ground frame $S$ is $v_{f} = 0 \text{ m s}^{-1}$.

Final velocity of the block in frame $S'$ is:
$v' = v_{f} - v_{F} = 0 - 2.0 = -2.0 \text{ m s}^{-1}$

Initial kinetic energy of the block in frame $S'$:
$K_{i}' = \dfrac{1}{2} m (u')^2 = 0 \text{ J}$

Final kinetic energy of the block in frame $S'$:
$K_{f}' = \dfrac{1}{2} m (v')^2 = \dfrac{1}{2} \times 100 \times (-2.0)^2 = 200 \text{ J}$

Increase in kinetic energy of the block as seen from the moving frame is:
$\Delta K' = K_{f}' - K_{i}' = 200 - 0 = 200 \text{ J}$