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A body of mass $0.15 \mathrm{~kg}$ moving with a velocity of $15 \mathrm{~ms}^{-1}$ comes to rest, when it hits a spring that is fixed at another end. If the force constant of the spring is $1500 \mathrm{Nm}^{-1}$, then the compression in the spring is
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The correct answer is:
$0.15 \mathrm{~m}$
Given, mass of the body, $m=0.15 \mathrm{~kg}$
Velocity of the body, $v=15 \mathrm{~ms}^{-1}$
Spring constant, $k=1500 \mathrm{Nm}^{-1}$
According to law of conservation of energy, the change in kinetic energy is equal to the change in potential energy, i.e.
$$
\begin{aligned}
\frac{1}{2} m v^2-0 & =0+\frac{1}{2} k x^2 \\
\Rightarrow \quad x^2 & =\frac{m}{k} v^2 \Rightarrow x=v \sqrt{\frac{m}{k}}=15 \times \sqrt{\frac{0.15}{1500}} \\
& =0.15 \mathrm{~m}
\end{aligned}
$$
Velocity of the body, $v=15 \mathrm{~ms}^{-1}$
Spring constant, $k=1500 \mathrm{Nm}^{-1}$
According to law of conservation of energy, the change in kinetic energy is equal to the change in potential energy, i.e.
$$
\begin{aligned}
\frac{1}{2} m v^2-0 & =0+\frac{1}{2} k x^2 \\
\Rightarrow \quad x^2 & =\frac{m}{k} v^2 \Rightarrow x=v \sqrt{\frac{m}{k}}=15 \times \sqrt{\frac{0.15}{1500}} \\
& =0.15 \mathrm{~m}
\end{aligned}
$$
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