Question

Find the coefficient of friction if the time taken by a block on the rough surface is \(50\%\) more than the time taken on a smooth surface. The distance slid by the mass is the same in both cases. The incline angle is \(45^\circ\).

• A. \( \mu = \frac{5}{7} \)
• B. \( \mu = \frac{4}{9} \)
• C. \( \mu = \frac{4}{7} \)
• D. \( \mu = \frac{5}{9} \)

Hint:

When distance travelled is the same for motions starting from rest, use \( s = \tfrac{1}{2}at^2 \). Thus \( a_1 t_1^2 = a_2 t_2^2 \). For an inclined plane with friction, acceleration is \( g(\sin\theta - \mu\cos\theta) \).

Answer 1

The Correct Option is D

Concept: For motion starting from rest with constant acceleration, \[ s = \frac{1}{2}at^2 \] If the distance is the same in two cases, then \[ a_1 t_1^2 = a_2 t_2^2 \] Acceleration on a smooth incline: \[ a_1 = g\sin\theta \] Acceleration on a rough incline: \[ a_2 = g(\sin\theta - \mu \cos\theta) \] Step 1: Write the relation between the times.} Time on rough surface is \(50\%\) more than smooth surface. \[ t_2 = 1.5\, t_1 \] \[ t_2^2 = 2.25\, t_1^2 \] Step 2: Use equal distance condition.} \[ a_1 t_1^2 = a_2 t_2^2 \] \[ a_1 = 2.25\, a_2 \] Step 3: Substitute accelerations.} \[ g\sin\theta = 2.25\, g(\sin\theta - \mu\cos\theta) \] Cancel \(g\): \[ \sin\theta = 2.25(\sin\theta - \mu\cos\theta) \] Step 4: Substitute \( \theta = 45^\circ \).} \[ \sin45^\circ = \cos45^\circ \] \[ 1 = 2.25(1 - \mu) \] \[ 1 = 2.25 - 2.25\mu \] \[ 2.25\mu = 1.25 \] \[ \mu = \frac{1.25}{2.25} \] \[ \mu = \frac{5}{9} \]