When a force is applied to a spring of force constant K, elongation produced in the spring is x. When the force is slowly increased to twice is present value, work done on the spring will be

When a force is applied to a spring of force constant K, elongation produced in the spring is x. When the force is slowly increased to twice is present value, work done on the spring will be

A
$\frac{1}{2} K x^{2}$
B
$K x^{2}$
C
$\frac{3 K x^{2}}{2}$
D
$2 K x^{2}$

Solution:

$F = K \cdot x, 2 F = K \cdot x^{1} \Rightarrow x^{1} = \frac{2 F}{K}$

$∴ Work done = \frac{1}{2} K {(x^{1})}^{2} - \frac{1}{2} {Kx}^{2} = \frac{1}{2} K {(\frac{2 F}{K})}^{2} - \frac{1}{2} K \cdot {(\frac{F}{K})}^{2}$

$= \frac{3 F^{2}}{2 K} = \frac{3 (Kx)^{2}}{2 K} = \frac{3}{2} {Kx}^{2}$

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