Question

A block slides down a \(30^\circ\) inclined plane. Coefficient of friction on upper half is \[ \mu_1=\frac{1}{2\sqrt3} \] and on lower half is \[ \mu_2=\frac{1}{4\sqrt3} \] Find ratio of velocities at midpoint and bottom.

• A. \(2:3\)
• B. \(\sqrt2:\sqrt5\)
• C. \(2:5\)
• D. \(\sqrt2:\sqrt3\)

Hint:

For inclined plane with changing friction, solve motion separately in each region and use continuity of velocity.

Answer 1

The Correct Option is B

Concept: Acceleration on inclined plane with friction: \[ a=g(\sin\theta-\mu\cos\theta) \] Velocity relation: \[ v^2=u^2+2as \]

Step 1: Find acceleration on upper half.
For upper half \[ a_1 = g \left( \sin30^\circ-\mu_1\cos30^\circ \right) \] \[ = g \left( \frac12-\frac{1}{2\sqrt3}\times\frac{\sqrt3}{2} \right) \] \[ = g \left( \frac12-\frac14 \right) \] \[ = \frac{g}{4} \]

Step 2: Velocity at midpoint.
Let total length be \(L\) Distance covered \[ \frac{L}{2} \] Initially at rest. \[ v_1^2 = 2a_1\frac{L}{2} \] \[ v_1^2 = \frac{gL}{4} \]

Step 3: Acceleration on lower half.
\[ a_2 = g \left( \frac12-\frac{1}{4\sqrt3}\times\frac{\sqrt3}{2} \right) \] \[ = g \left( \frac12-\frac18 \right) \] \[ = \frac{3g}{8} \]

Step 4: Velocity at bottom.
Using second half motion \[ v_2^2 = v_1^2+2a_2\frac{L}{2} \] \[ = \frac{gL}{4} + \frac{3gL}{8} \] \[ = \frac{5gL}{8} \] Thus ratio \[ \frac{v_1}{v_2} = \sqrt{ \frac{gL/4}{5gL/8} } \] \[ = \sqrt{\frac25} \] Hence \[ v_1:v_2 = \boxed{\sqrt2:\sqrt5} \]