1. Analysis of the System

The plank moves with a constant velocity $u$. When the block of mass $m$ is placed on it, there is relative motion between them. Kinetic friction $f_k$ acts forward on the block (accelerating it) and backward on the plank (opposing its motion).

Since the plank maintains a constant velocity $u$, the external agency pushing the plank must exert a force $F_{agency}$ exactly equal and opposite to the frictional force exerted by the block on the plank.

$$ F_{agency} = f_k = \mu m g $$

2. Work Done by the Agency

The block accelerates from rest to velocity $u$. The time $t$ taken for this can be found using the impulse-momentum equation for the block or kinematic equations.

Acceleration of block: $a = \frac{f_k}{m} = \mu g$

Time to reach velocity $u$: $t = \frac{u}{a} = \frac{u}{\mu g}$

During this time $t$, the plank moves a distance $S_p$:

$$ S_p = u \times t = u \left( \frac{u}{\mu g} \right) = \frac{u^2}{\mu g} $$

The work done by the agency ($W$) is force times displacement:

$$ W = F_{agency} \cdot S_p = (\mu m g) \cdot \left( \frac{u^2}{\mu g} \right) = m u^2 $$

3. Calculation

We are given:

• Work done, $W = 11.25$ J
• Velocity, $u = 1.5$ m/s

Rearranging the derived formula $W = m u^2$ to solve for mass $m$:

$$ m = \frac{W}{u^2} $$ $$ m = \frac{11.25}{(1.5)^2} = \frac{11.25}{2.25} $$ $$ m = 5.0 \text{ kg} $$

Alternative Approach (Work-Energy Theorem)

Work done by agency = Change in Kinetic Energy of system + Heat generated due to friction.
$\Delta K_{plank} = 0$ (constant velocity)
$\Delta K_{block} = \frac{1}{2} m u^2$
Heat $Q = f_k \times S_{relative} = (\mu m g) \times \frac{u^2}{2\mu g} = \frac{1}{2} m u^2$
$W = \frac{1}{2} m u^2 + \frac{1}{2} m u^2 = m u^2$