Question

A block of mass 1.5 kg kept on a rough horizontal surface is given a horizontal velocity of 10 ms\(^{-1}\). If the block comes to rest after travelling a distance of 12.5 m, the coefficient of kinetic friction between the surface and the block is (Acceleration due to gravity = 10 ms\(^{-2}\))

• A. 0.2
• B. 0.4
• C. 0.8
• D. 0.6

Hint:

In problems involving kinetic friction, use the work-energy theorem to relate the loss in kinetic energy to the work done by the friction force.

Answer 1

The Correct Option is B

Step 1: We can use the work-energy principle to solve this problem. The work done by the friction force will be equal to the loss in kinetic energy of the block. The equation for kinetic energy is: \[ KE = \frac{1}{2} m v^2 \] where \( m = 1.5 \, \text{kg} \) and \( v = 10 \, \text{m/s} \). Thus, the initial kinetic energy is: \[ KE = \frac{1}{2} \times 1.5 \times (10)^2 = 75 \, \text{J} \] Step 2: The work done by the friction force \( W_f \) is given by: \[ W_f = F_f \times d = \mu mg \times d \] where \( \mu \) is the coefficient of kinetic friction, \( m = 1.5 \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \), and \( d = 12.5 \, \text{m} \). \[ W_f = \mu \times 1.5 \times 10 \times 12.5 = 187.5 \mu \, \text{J} \] Step 3: Since the block comes to rest, the work done by the friction force is equal to the initial kinetic energy: \[ 187.5 \mu = 75 \] \[ \mu = \frac{75}{187.5} = 0.4 \] Thus, the coefficient of kinetic friction is \( \mu = 0.4 \).