Question

A block of mass $m$ is hung from a pulley of mass $M$ and radius $R$ and then released from rest. Initially the block is at height $h$ from the floor. The speed of the block when it strikes the floor if $M=4m$ will be

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

Answer 1

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

The initial gravitational potential energy of the block is transformed into translational kinetic energy of the block and rotational kinetic energy of the pulley.

If $v$ is the speed of the block just before it hits the floor, then from conservation of energy, we have

$K.E_i+P.E_i=K.E_f+P.E_f$

$\Rightarrow 0+mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+0$

$\Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\times \left(\frac{1}{2}M{R}^2\right)\times {\left(\frac{v}{R}\right)}^2$

$\Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{4}M{v}^2$

$\Rightarrow v=\sqrt{\frac{mgh}{(\frac{m}{2}+\frac{M}{4})}}$

Putting $M=4m$, we get

$v=\sqrt{\frac{2gh}{3}}$

Hence, option $\left(C\right)$ is correct.