Question

The figure shows a block of mass $M$ resting on a horizontal surface. The acceleration with which a boy of mass $m$ should climb the rope (neglect mass) so as to lift the block should be greater than:

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

Answer 1

Answer: (B)

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

The tension lifts the boys weight, and also accelerates him.
Therefore T $=$ m(g+a).
In the limiting case, the big block does not move. Therefore T$=$Mg.
hence m(g+a)$=$ Mg for the limiting case.
hence $a=(\frac{M}{m}-1)g$