Question

A block of mass $5.0\text{kg}$ slides down from the top of an inclined plane of length $3\text{m}$. The first $1\text{m}$ of the plane is smooth and the next $2\text{m}$ 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. $\frac{2}{\sqrt{2}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{\sqrt{3}}{4}$
• D. $\frac{\sqrt{3}}{5}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$

FBD of the block at inital and final positions:


From work energy theorem, $W_{ext}=\Delta U+\Delta K.E$
Since the only external non conservative force is friction and $\Delta K.E=0$
we have $\Delta U=W_f$
i.e Loss in gravitational potential energy of the block = Work done against friction,
$\left(mg\right)\times \left(3\sin {30}^{\circ }\right)=\left(\mu mg\cos \theta \right)\times 2$
$\Rightarrow 3\times \frac{1}{2}=\mu \times \frac{\sqrt{3}}{2}\times 2$
$\Rightarrow \mu =\frac{\sqrt{3}}{2}$