Question

A block of mass M is tied to one end 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 wedge 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 wedge when the block meets the pulley? (man does not leave his position during the pull) (on the block 2M.)

• A. 0.5 m
• B. 1 m
• C. ZERO
• D. 2/3 m

Answer 1

The correct option is A 0.5 m

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 of the block

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

So, deceleration, $a=\frac{F}{m}=\frac{\mu mg}{m}=\mu g$

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 $4m\times a_1=\mu mg$

$a_1=\frac{\mu g}{4}$

Again applying the equation of motion

$S_1=ut+\frac{1}{2}{a}_1{t}^2=0+\frac{1}{2}\times \frac{\mu g}{4}\times \frac{4}{\mu g}=0.5m$