Question

A block of mass \( M \) is moving on rough horizontal surface with momentum \( P \). The coefficient of friction between the block and surface is \( \mu \). The distance covered by the block before it stops is

• A. \( \frac{2 \mu M g}{P} \)
• B. \( \frac{P}{2 \mu M g} \)
• C. \( \frac{P^2}{2 \mu M g} \)
• D. \( \frac{2 \mu M^2 g}{P^2} \)

Hint:

The work done by friction equals the change in kinetic energy. Use this principle to solve for the distance traveled before an object stops.

Answer 1

The Correct Option is C

Step 1: Work-Energy Principle.
The work done by the frictional force \( F_f = \mu M g \) is equal to the change in kinetic energy of the block. The initial kinetic energy is \( \frac{P^2}{2M} \), and the final kinetic energy is 0 (since the block stops). The work done by friction is \( W = F_f \times d = \mu M g \times d \), where \( d \) is the distance.
Step 2: Applying the Work-Energy Theorem.
According to the work-energy theorem: \[ W = \Delta K = 0 - \frac{P^2}{2M} \] Thus: \[ \mu M g \times d = \frac{P^2}{2M} \] Solving for \( d \): \[ d = \frac{P^2}{2 \mu M g} \]
Step 3: Conclusion.
The distance covered by the block before it stops is \( \frac{P^2}{2 \mu M g} \), so the correct answer is (C).