Question

A block of mass 5.0 kg slides down from the top of an inclined plane of length 3 m. The first 1 m of the plane is smooth and the next 2 m is rough. The block is released from rest and again comes to rest at the bottom of the plane. If the plane is inclined at ${30}^{\circ }$ with the horizontal, find the coefficient of friction on the rough portion.

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{\sqrt{3}}{4}$
• D. $\frac{\sqrt{3}}{5}$

Answer 1

The correct option is B

$\frac{\sqrt{3}}{2}$

Force Method:

Form P to Q: $v^2=O^2+2a_1{S}_1$

Where $a_1=gsin{30}^{\circ },s_1=1m$

And form Q to R: $O^2=v^2+2a_2{s}_2$

Where $a_2=gsin{30}^{\circ }-\mu gcos{30}^{\circ },s_2=2m$

Solve to get $\mu =\frac{\sqrt{3}}{2}$