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An elastic spring of wnstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done dwring the second stretching is
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Verified Answer
The correct answer is:
$\frac{k y}{2}(2 x+y)$
According to the question, In the first condition,
$$
W_1=\frac{1}{2} k x^2
$$
In the second condition,
$$
W_2=\frac{1}{2} k(x+y)^2
$$
Work for $x$ to $y$ expansion
$$
\begin{aligned}
& =\frac{1}{2} k(x+y)^2-\frac{1}{2} k x^2 \\
& =\frac{1}{2} k\left[x^2+y^2+2 x y-x^2\right] \\
& =\frac{1}{2} k\left[y^2+2 x y\right] \\
& =\frac{1}{2} k y[y+2 x]
\end{aligned}
$$
$$
W_1=\frac{1}{2} k x^2
$$
In the second condition,
$$
W_2=\frac{1}{2} k(x+y)^2
$$
Work for $x$ to $y$ expansion
$$
\begin{aligned}
& =\frac{1}{2} k(x+y)^2-\frac{1}{2} k x^2 \\
& =\frac{1}{2} k\left[x^2+y^2+2 x y-x^2\right] \\
& =\frac{1}{2} k\left[y^2+2 x y\right] \\
& =\frac{1}{2} k y[y+2 x]
\end{aligned}
$$
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