Question

In $Fig.6.331$, the block of mass M is at rest on the floor. The acceleration with which a boy of mass m should climb along the rope of negligible mass so as to lift the block from the floor is:

• A. $(\frac{2M}{m}-1)g$
• B. $(\frac{M}{m}-1)g$
• C. $\frac{M}{m}g$
• D. $>\frac{M}{m}g$

Answer 1

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

Let T be the tension in the string. The free body diagram for the boy of mass m is,(refer image 1)

The net force on the boy is,

$T-mg=ma$ ...........(1)

The free body diagram for the block of mass M is,(refer image 2)

For the block to be lifted fro the tension in the string exceed the weight of the block.

$T=Mg$ ...........(2)

Substitute the value of T from (2) to (1)

$T-mg=ma$

$Mg-mg=ma$

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