Question

As shown in the figure, M is a man of mass 60 kg standing on a block of mass 40 kg kept on ground. The co-efficient of friction between the feet of the man and the block is 0.3 and that between B and the ground is 0.1. If the man accelerates with an acceleration $2m/s^2$ in the forward direction, then:

• A. it is not possible.
• B. B will move backwards with an acceleration $0.5m/s^2$.
• C. B will not move.
• D. B will move forward with an acceleration $0.5m/s^2$.

Answer 1

The correct option is B B will move backwards with an acceleration $0.5m/s^2$.

Let us begin this problem with an assumption that the block will remain stationary.

The force on the man is: $F=m{a}_{man}=\left(60kg\right)\left(2m/s^2\right)=120N$

towards left.

Now, the force required for the block to remain stationary is $120N$ towards left on block from ground.

But since normal force on block from ground is $N=(40+60kg)\left(10m/s^2\right)=1000N$, maximum friction on it is $f=\left(1000N\right)\left(0.1\right)=100N$ towards left (since block has a tendancy to move towards right).

Hence, block will move towards left and force on it will be $120-100N=20N$.

Therefore, acceleration of the block is $a_{block}=20N/40kg=0.5m/s^2$ towards left, i.e. backwards.