Question

Two blocks \(A\) and \(B\) are released from rest on two inclined planes of the same inclination. One inclined plane is smooth while the other is rough as shown in the figure. If block \(B\) takes \(50\%\) more time to reach the bottom than block \(A\), find the coefficient of friction \((\mu)\). The inclination of the plane is \(45^\circ\).

• A. \( \frac{5}{9} \)
• B. \( \frac{13}{9} \)
• C. \( \frac{3}{9} \)
• D. \( \frac{7}{9} \)

Answer 1

The Correct Option is A

Concept:

For motion down an inclined plane: Acceleration on a smooth incline \[ a = g\sin\theta \] Acceleration on a rough incline \[ a = g(\sin\theta - \mu\cos\theta) \] If a body starts from rest and moves a distance \(S\), \[ S = \frac{1}{2}at^2 \] Thus, \[ t = \sqrt{\frac{2S}{a}} \] Hence time is inversely proportional to the square root of acceleration. \[ t \propto \frac{1}{\sqrt{a}} \] Step 1: Acceleration of block \(A\) (smooth plane). \[ a_A = g\sin45^\circ \] \[ a_A = \frac{g}{\sqrt{2}} \] Step 2: Acceleration of block \(B\) (rough plane). \[ a_B = g(\sin45^\circ - \mu\cos45^\circ) \] \[ a_B = g\left(\frac{1}{\sqrt{2}} - \mu\frac{1}{\sqrt{2}}\right) \] \[ a_B = \frac{g}{\sqrt{2}}(1-\mu) \] Step 3: Using the relation between time and acceleration. \[ t \propto \frac{1}{\sqrt{a}} \] \[ \frac{t_B}{t_A} = \sqrt{\frac{a_A}{a_B}} \] Given \(B\) takes \(50\%\) more time: \[ t_B = 1.5\, t_A \] \[ \frac{t_B}{t_A} = \frac{3}{2} \] Step 4: Substitute accelerations. \[ \frac{3}{2} = \sqrt{\frac{\frac{g}{\sqrt{2}}}{\frac{g}{\sqrt{2}}(1-\mu)}} \] \[ \frac{3}{2} = \sqrt{\frac{1}{1-\mu}} \] Step 5: Solve for \(\mu\). \[ \left(\frac{3}{2}\right)^2 = \frac{1}{1-\mu} \] \[ \frac{9}{4} = \frac{1}{1-\mu} \] \[ 1-\mu = \frac{4}{9} \] \[ \mu = \frac{5}{9} \]