Question

A block of mass $10kg$ is placed on a horizontal frictionless surface and is attached to a cord which passes over two light frictionless pulleys as shown in the figure. The hanging block tied to the other end of the cord is initially at rest $2m$ above the horizontal floor. If the hanging block strikes the floor $2s$ after the system is released, the weight of the hanging block is________($g=10m{s}^{-2}$)

• A. 22.22 N
• B. 11.11N
• C. 1.11 N
• D. 2.22 N

Answer 1

Answer: (B)

For hanging block,
$s=ut+\frac{1}{2}a{t}^2[\because u=0]$
$\Rightarrow 2=\frac{1}{2}\times a\times 2^2$
$\Rightarrow a=1m{s}^{-2}$
So, velocity of hanging block $=$ velocity of $10kg$ block after $2s$
$\Rightarrow v=u+at$
$\Rightarrow v=0+1\times 2$ or $v=2m{s}^{-1}$
From work - energy theorem,
KE of $10kg$ block $=$ Work done by gravity on hanging block
$=\frac{1}{2}\times 10\times 2^2=m\times 10\times 2$
$\Rightarrow m=1kg$
Weight of hanging block $=1\times 10=10kg\approx 11.11N$