Work done in stretching a spring of constant k by x is \(\frac{1}{2} \ kx^2\).The work done when above spring is stretched further from x to 2x is

CONCEPT:

• When we exert tensile stress on a wire, it will get stretched and work done in stretching the wire will be equal and opposite to the work done by inter-atomic restoring force. This work stored in the wire in the form of Elastic potential energy.

• The work done by an external force on spring is given by

\(\Rightarrow W = \frac{1}{2} k x^{2}\)

Where K = Force constant, x = Displacement of the spring

• The work  done when s spring stretched from x₁ to x₂ is given 

\(\Rightarrow W = \frac{K}{2} (x_{2}^{2} - x^{2}_{1})\)

CALCULATION :

Here x₂ = 2x ,x₁ = x

• The work done to stretch is given by

\(\Rightarrow W = \frac{K}{2} ((2x)^{2} - x^{2})\)

\(\Rightarrow W = \frac{K}{2} (4x^{2} - x^{2})\)

\(\Rightarrow W = \frac{3}{2}K x^{2}\)

• Hence, option 3 is the answer