An elastic spring of unstretched length L and | vedclass

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(D) The work done in stretching a spring by an extension $x$ is given by $W = \frac{1}{2} k x^2$. Initially,the spring is stretched by $x$. The potent

Vedclass · Accepted Answer

$\frac{k y}{2}(2 x+y)$

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(D) The work done in stretching a spring by an extension $x$ is given by $W = \frac{1}{2} k x^2$. Initially,the spring is stretched by $x$. The potential energy stored is $U_1 = \frac{1}{2} k x^2$. Then,it is further stretched by $y$,so the total extension becomes $(x + y)$. The potential energy stored is $U_2 = \frac{1}{2} k (x + y)^2$. The work done during the second stretching is the change in potential energy: $W = U_2 - U_1 = \frac{1}{2} k (x + y)^2 - \frac{1}{2} k x^2$. Expanding the term $(x + y)^2$: $W = \frac{1}{2} k (x^2 + y^2 + 2xy - x^2) = \frac{1}{2} k (y^2 + 2xy)$. Factoring out $y$: $W = \frac{ky}{2} (2x + y)$.

An elastic spring of unstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done during the second stretching is

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