Question

A block of mass $10\, kg$ 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 $2\, m$ above the horizontal floor. If the hanging block strikes the floor $2\, s$ after the system is released, the weight of the hanging block is________($g = 10\, ms^{-2}$)

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

Answer 1

Answer: (B)

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