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

Let the initial velocity of the block be in the positive $x$-direction. The ground frame is $S$ and the moving frame is $S'$, which moves with velocity $V = 2.0 \text{ m s}^{-1}$ in the positive $x$-direction.

In the frame $S'$, the initial velocity of the block is $v'_0 = 2.0 - 2.0 = 0 \text{ m s}^{-1}$. The belt moves with a constant velocity $v'_b = 0 - 2.0 = -2.0 \text{ m s}^{-1}$.

The block experiences a kinetic friction force $f_k = \mu_k m g$ in the negative $x$-direction (the direction of the belt's motion relative to the block). The acceleration of the block in $S'$ is $a' = -\dfrac{f_k}{m}$.

The time $t$ taken for the block to reach the velocity of the belt is obtained from $v'_b = v'_0 + a' t$, giving $-V = 0 - \dfrac{f_k}{m} t \Rightarrow t = \dfrac{m V}{f_k}$.

During this time $t$, the displacement of the belt in $S'$ is $d'_b = v'_b t = -V \left(\dfrac{m V}{f_k}\right) = -\dfrac{m V^2}{f_k}$.

To keep the belt fixed in the ground frame, an external force $F_{ext}$ must be applied to balance the friction force exerted by the block on the belt. By Newton's third law, the friction on the belt is in the positive $x$-direction, so $F_{ext} = -f_k$.

The work done by the external force in the moving frame $S'$ is $W'_{ext} = F_{ext} d'_b = (-f_k) \left(-\dfrac{m V^2}{f_k}\right) = m V^2$.

Substituting $m = 100 \text{ kg}$ and $V = 2.0 \text{ m s}^{-1}$:

$W'_{ext} = 100 \times (2.0)^2 = 400 \text{ J}$.

Answer: $400 \text{ J}$