Question

A plank is moving along a smooth surface with a constant speed V. A block of mass M is gently placed on it. Initially the blocks slips and then acquires the constant speed V same as the plank. Through the period, a horizontal force is applied on the plank to keep its speed constant.

• A. The work done by all the frictional force on block will be positive.
• B. Net work done by all the force on plank will be zero.
• C. The work performed by the external force Mv².
• D. Heat developed due to friction between the block and the plank $\frac{1}{2}Mv^2$.

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The problem describes a scenario where a block is placed on a moving plank, and the plank's speed is maintained constant by an external force. We need to analyze the work done by various forces and the heat generated.

Let:

• M be the mass of the block.
• V be the constant speed of the plank.
• t be the time taken for the block to acquire the speed V.
• f_k be the magnitude of the kinetic friction force between the block and the plank.

1. Motion of the Block:

Initially, the block is at rest ($u_b = 0$). It accelerates due to kinetic friction from the plank and eventually reaches a speed V.

Let a_b be the acceleration of the block.

From Newton's second law for the block: $f_k = M a_b$.

Using the first equation of motion: $v_b = u_b + a_b t \implies V = 0 + a_b t \implies a_b = \frac{V}{t}$.

So, the friction force is $f_k = M \frac{V}{t}$.

The displacement of the block during time t is $S_b = u_b t + \frac{1}{2} a_b t^2 = 0 + \frac{1}{2} \left(\frac{V}{t}\right) t^2 = \frac{1}{2} V t$.

2. Motion of the Plank:

The plank moves with a constant speed V. This means its acceleration is zero.

Forces acting on the plank in the horizontal direction:

• External force $F_{ext}$ (forward, to maintain constant speed).
• Kinetic friction force $f_k'$ exerted by the block on the plank (backward, opposing the plank's motion relative to the block).

By Newton's third law, $f_k' = f_k$.

Since the plank's acceleration is zero, the net force on it is zero: $F_{ext} - f_k = 0 \implies F_{ext} = f_k$.

So, $F_{ext} = M \frac{V}{t}$.

The displacement of the plank during time t is $S_p = V t$.

Now, let's evaluate each option:

(A) The work done by all the frictional force on block will be positive.

The kinetic friction force $f_k$ acts on the block in the forward direction (the direction of motion of the plank, which causes the block to accelerate). The block also moves in the forward direction. Since the force and displacement are in the same direction, the work done by friction on the block is positive.

Work done by friction on block $W_{f,b} = f_k \times S_b = \left(M \frac{V}{t}\right) \times \left(\frac{1}{2} V t\right) = \frac{1}{2} M V^2$.

This value is positive.

Thus, option (A) is correct.

(B) Net work done by all the force on plank will be zero.

The plank moves with a constant speed V. This means its kinetic energy does not change ($\Delta KE_{plank} = 0$). According to the Work-Energy Theorem, the net work done on an object is equal to the change in its kinetic energy.

$W_{net,plank} = \Delta KE_{plank} = 0$.

Thus, option (B) is correct.

(C) The work performed by the external force Mv².

The work done by the external force is $W_{ext} = F_{ext} \times S_p$.

We found $F_{ext} = M \frac{V}{t}$ and $S_p = V t$.

$W_{ext} = \left(M \frac{V}{t}\right) \times (V t) = M V^2$.

Thus, option (C) is correct.

(D) Heat developed due to friction between the block and the plank $\frac{1}{2}Mv^2$.

The heat developed due to friction is equal to the product of the friction force and the relative displacement between the surfaces.

The relative displacement $S_{rel} = S_p - S_b$.

$S_{rel} = V t - \frac{1}{2} V t = \frac{1}{2} V t$.

Heat developed $Q = f_k \times S_{rel} = \left(M \frac{V}{t}\right) \times \left(\frac{1}{2} V t\right) = \frac{1}{2} M V^2$.

Alternatively, using the work-energy theorem for the system (block + plank):

Work done by external force = Change in kinetic energy of the system + Heat developed.

$W_{ext} = \Delta KE_{system} + Q$.

Initial kinetic energy of the system $KE_{initial} = \frac{1}{2} M_{plank} V^2$ (assuming $M_{plank}$ is the mass of the plank).

Final kinetic energy of the system $KE_{final} = \frac{1}{2} M_{plank} V^2 + \frac{1}{2} M V^2$.

Change in kinetic energy of the system $\Delta KE_{system} = KE_{final} - KE_{initial} = \frac{1}{2} M V^2$.

From option (C), $W_{ext} = M V^2$.

Substituting these values into the equation:

$M V^2 = \frac{1}{2} M V^2 + Q$.

$Q = M V^2 - \frac{1}{2} M V^2 = \frac{1}{2} M V^2$.

Thus, option (D) is correct.

All the given options (A), (B), (C), and (D) are correct.