1. Determine Final Velocity

A block of mass $m$ is given an initial velocity $u$ on a plank of mass $M$. The system rests on a frictionless floor. Friction $\mu$ acts between the block and the plank.

Momentum is conserved for the system because there are no external horizontal forces.

$$p_i = p_f$$

$$m u + M(0) = (m + M)v_f$$

The final common velocity $v_f$ is:

$$v_f = \frac{mu}{m + M}$$

2. Calculate Total Heat Dissipated (Work Done)

The thermal energy dissipated is equal to the loss in Kinetic Energy of the system.

$$Q = K_i – K_f$$

$$Q = \frac{1}{2}mu^2 – \frac{1}{2}(m+M)v_f^2$$

Substituting $v_f$:

$$Q = \frac{1}{2}mu^2 – \frac{1}{2}(m+M)\left(\frac{mu}{m+M}\right)^2$$

$$Q = \frac{1}{2}mu^2 – \frac{1}{2}\frac{m^2 u^2}{(m+M)}$$

$$Q = \frac{1}{2}mu^2 \left( 1 – \frac{m}{m+M} \right) = \frac{1}{2}mu^2 \left( \frac{M}{m+M} \right) = \frac{mM u^2}{2(m+M)}$$

3. Calculate the Time Taken to Stop Sliding

We need the time $\Delta t$ for the relative velocity to become zero. We can analyze the forces:

• Force on block $m$: $f = -\mu mg$ (retardation) $\implies a_m = -\mu g$
• Force on plank $M$: $f = +\mu mg$ (acceleration) $\implies a_M = \frac{\mu mg}{M}$

The relative acceleration $a_{rel}$ (block relative to plank) is:

$$a_{rel} = a_m – a_M = -\mu g – \frac{\mu mg}{M} = -\mu g \left( \frac{M+m}{M} \right)$$

Using the kinematic equation for relative motion ($v_{rel} = u_{rel} + a_{rel}t$), where final relative velocity is 0 and initial is $u$:

$$0 = u – \left[ \mu g \left( \frac{m+M}{M} \right) \right] t$$

$$t = \frac{u}{\mu g \left( \frac{m+M}{M} \right)} = \frac{Mu}{\mu g (m+M)}$$

4. Calculate Average Power

Average power dissipated is the total heat energy divided by the time taken.

$$P_{avg} = \frac{Q}{t}$$

Substitute the expressions derived in Step 2 and Step 3:

$$P_{avg} = \frac{\left[ \frac{mM u^2}{2(m+M)} \right]}{\left[ \frac{Mu}{\mu g (m+M)} \right]}$$

Rearranging the terms:

$$P_{avg} = \frac{mM u^2}{2(m+M)} \cdot \frac{\mu g (m+M)}{Mu}$$

Canceling common terms $(m+M)$, $M$, and one $u$:

$$P_{avg} = \frac{1}{2} \mu m g u = \frac{\mu m g u}{2}$$