Question

A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is (Take \( g = 10 \, \text{m/s}^2 \))

• A. 0.06
• B. 0.03
• C. 0.04
• D. 0.01

Hint:

To calculate the coefficient of friction, use the equation of motion and the relationship between frictional force and normal force.

Answer 1

The Correct Option is C

Step 1: Using the Equation of Motion.
The frictional force \( f = \mu N \), where \( N = mg \) is the normal force. Using the equation of motion for stopping: \[ v = u + at \quad \Rightarrow \quad 0 = 6 + a \times 10 \] Solving for acceleration \( a = -0.6 \, \text{m/s}^2 \). The frictional force is \( f = ma = 2 \times 0.6 = 1.2 \, \text{N} \). The coefficient of friction is: \[ \mu = \frac{f}{N} = \frac{1.2}{2 \times 10} = 0.04 \] Step 2: Conclusion.
The correct answer is (C), 0.04.