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

Step 1: Write the acceleration on an incline with friction.
$a=g(\sin\theta-\mu\cos\theta)$ with $\theta=30^\circ$, so $\sin30^\circ=\dfrac12$ and $\cos30^\circ=\dfrac{\sqrt3}{2}$.
Step 2: Upper half acceleration.
$a_1=g\left(\dfrac12-\dfrac{1}{2\sqrt3}\cdot\dfrac{\sqrt3}{2}\right)=g\left(\dfrac12-\dfrac14\right)=\dfrac{g}{4}$.
Step 3: Speed at the midpoint.
Starting from rest over distance $\dfrac{L}{2}$, $v_1^2=2a_1\cdot\dfrac{L}{2}=a_1L=\dfrac{gL}{4}$.
Step 4: Lower half acceleration.
$a_2=g\left(\dfrac12-\dfrac{1}{4\sqrt3}\cdot\dfrac{\sqrt3}{2}\right)=g\left(\dfrac12-\dfrac18\right)=\dfrac{3g}{8}$.
Step 5: Speed at the bottom.
On the lower half, $v_2^2=v_1^2+2a_2\cdot\dfrac{L}{2}=\dfrac{gL}{4}+\dfrac{3gL}{8}=\dfrac{5gL}{8}$.
Step 6: Take the ratio.
$\dfrac{v_1^2}{v_2^2}=\dfrac{gL/4}{5gL/8}=\dfrac{2}{5}$, so $v_1:v_2=\sqrt2:\sqrt5$.
\[ \boxed{\sqrt2:\sqrt5} \]