Correct Answer - Option 3 :
\(\frac{3}{2} \ kx^2\)
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