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A block of mass $10 \mathrm{~kg}$ 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 $2 \mathrm{~m}$ above the horizontal floor.If the hanging block strikes the floor $2 \mathrm{~s}$ after the system is released, then weight of the hanging block is ....... $\left(g=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
11.11 N
For hanging block,
$$
\begin{aligned}
& \quad s=u t+\frac{1}{2} a t^2[\because u=0] \Rightarrow 2=\frac{1}{2} \times a \times 2^2 \\
& \Rightarrow \quad a=1 \mathrm{~ms}^{-2}
\end{aligned}
$$
So, velocity of hanging block $=$ velocity of $10 \mathrm{~kg}$ block after $2 \mathrm{~s}$
$$
\begin{array}{ll}
\Rightarrow & v=u+a t \\
\Rightarrow & v=0+1 \times 2 \text { or } v=2 \mathrm{~ms}^{-1}
\end{array}
$$
From work - energy theorem,
$\mathrm{KE}$ of $10 \mathrm{~kg}$ block $=$ Work done by gravity on hanging block $=\frac{1}{2} \times 10 \times 2^2=m \times 10 \times 2$
$$
\Rightarrow \quad m=1 \mathrm{~kg}
$$
Weight of hanging block $=1 \times 10=10 \mathrm{~kg} \approx 11.1 \mathrm{lN}$
$$
\begin{aligned}
& \quad s=u t+\frac{1}{2} a t^2[\because u=0] \Rightarrow 2=\frac{1}{2} \times a \times 2^2 \\
& \Rightarrow \quad a=1 \mathrm{~ms}^{-2}
\end{aligned}
$$
So, velocity of hanging block $=$ velocity of $10 \mathrm{~kg}$ block after $2 \mathrm{~s}$
$$
\begin{array}{ll}
\Rightarrow & v=u+a t \\
\Rightarrow & v=0+1 \times 2 \text { or } v=2 \mathrm{~ms}^{-1}
\end{array}
$$
From work - energy theorem,
$\mathrm{KE}$ of $10 \mathrm{~kg}$ block $=$ Work done by gravity on hanging block $=\frac{1}{2} \times 10 \times 2^2=m \times 10 \times 2$
$$
\Rightarrow \quad m=1 \mathrm{~kg}
$$
Weight of hanging block $=1 \times 10=10 \mathrm{~kg} \approx 11.1 \mathrm{lN}$
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