Question

In the figure the block of mass $M$ is at rest on the floor. The acceleration with which should a boy of mass $m$ climb along the rope of negligible mass so as to lift the block from the floor, is :

• A. $=(\frac{M}{m}-1)g$
• B. $>(\frac{M}{m}-1)g$
• C. $=\frac{M}{m}g$
• D. $>\frac{M}{m}g$

Answer 1

The correct option is B $>(\frac{M}{m}-1)g$

Equation of motion for $M$:

For just lift the block.

$\Rightarrow T=Mg$ $...\left(1\right)$

Since the boy moves up with an acceleration ${}^{\prime }{a}^{\prime }$,

$T-mg=ma$

$\Rightarrow T=m(g+a)$ $...\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$, we obtain

$Mg=m(g+a)$

$\Rightarrow a=(\frac{M}{m}-1)g$

That means, if $a>(\frac{M}{m}-1)g$, the block $M$ can be lifted.

Therefore $\left(B\right)$