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A spring with spring constant $k$ is extended from $x=0_{\mathrm{x}=0 \text { to } \mathrm{x}=\mathrm{x}_1} x=x_1$. The work done will be
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Verified Answer
The correct answer is:
$\frac{1}{2} k x_1^2$
we know that
work done = change in Energy
$\mathrm{W}=\Delta \mathrm{E}$
and energy of spring is defined as $\frac{1}{2} k x^2$ where $x$ is stretch in spring.
$\therefore \mathrm{W}=\frac{1}{2} \mathrm{kx}_1^2$
work done = change in Energy
$\mathrm{W}=\Delta \mathrm{E}$
and energy of spring is defined as $\frac{1}{2} k x^2$ where $x$ is stretch in spring.
$\therefore \mathrm{W}=\frac{1}{2} \mathrm{kx}_1^2$
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