Question

A spring of force constant 'k' is stretched by a small length 'x'. Find the work done in stretching it further by a small length 'y'.​

Answer 1

AnSwer :

Let W₁ is the work done in stretching a spring of froce constant 'K' through a length "X" Then,

${\sf {W_{1} = \dfrac{1}{2} {Kx}^{2}}} \: \: \: \: .....(1)$

Let "W₂" is the work done in stretching the spring a length (x + y) .Then

${\sf {W_{2} = \dfrac{1}{2} K {(x + y)}^{2} \: \: \: .....(2) }}$

∴ Additional work done, to increase the elongation by "y" is W = W₂ - W₁

using (1) and (2)

${ \sf {W = \dfrac{1}{2} K {(x + y)}^{2} - \dfrac{1}{2} K {x}^{2} }}$

$\dag \boxed{ \sf W = \frac{1}{2} Ky(y + 2x)}$

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