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  • 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. The work done in the second stretching

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. The work done in the second stretching is

Option: 1

\mathrm{\frac{1}{2} k y^2}


Option: 2

\mathrm{\frac{1}{2} k\left(x^2+y^2\right)}


Option: 3

\mathrm{\frac{1}{2} k(x+y)^2}


Option: 4

\mathrm{\frac{1}{2} k y(2 x+y)}


Answers (1)

best_answer

Potential energy stored in the spring when it is extended by x is

                             \mathrm{ U_1=\frac{1}{2} k x^2 }
Potential energy stored in the spring when it is further extended by y is

\mathrm{ \begin{aligned} U_2 & =\frac{1}{2} k(x+y)^2 \\\\ \therefore \text { Work done }=U_2-U_1 & =\frac{1}{2} k(x+y)^2-\frac{1}{2} k x^2 \\\\ & =\frac{1}{2} k y(2 x+y) \end{aligned} }
Hence the correct choice is (d).

Posted by

himanshu.meshram

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