Question

A block of mass $M$ is tied to one of a massless rope. the other end of the rope is in the hands of a man of mass $2M$ as shown in the figure. The block and the man are resting on a rough plank of mass $M$ as shown in the figure. The whole system is resting on a smooth horizontal surface. The man pulls the rope. Pulley is massless and frictionless. What is the displacement of the plank when the block meets the pulley?

(Man does not leave his position on plank during the pull)

• A. $0.5m$
• B. $1m$
• C. $0$
• D. $\frac{2}{3}m$

Answer 1

The correct option is A $0.5m$

When we consider the free body diagram of the small block we can see that the block will 'just' start moving when the pulling force of man overcomes the frictional force and we assume that force is sufficient to reach the block till the pulley.

$F\left(N\right)$ is the normal force on the block

frictional force $F=\mu \times F\left(N\right)=\mu \times mg=\mu mg$

so deceleration is $a=\frac{F}{m}=\frac{\mu mg}{m}=\mu g$

Using Newtons equations of motion, we know that the block's initial velocity is zero.

$S=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times \mu g\times t^2$

$t=\frac{2}{\sqrt{\mu g}}$ as $S=2m$, given

Now consider the bigger block, the force which is opposing the smaller block from moving is also acting on the bigger block to move forward

so $F=\mu mg$

so acceleration is $4m\times a^{{}^{\prime }}=\mu mg$

$a^{{}^{\prime }}=\frac{\mu g}{4}$

Again applying the equations of motion:

$S^{\prime }=ut+\frac{1}{2}{a}^{{}^{\prime }}{t}^2=0+\frac{1}{2}\times \frac{\mu g}{4}\times \frac{4}{\mu g}=0.5m$