Question

The spring extends by $x$ on loading then the energy stored by the spring is (if $T$ is the tension in spring and $K$ is spring constant)

• A. $\frac{T^2}{2K}$
• B. $\frac{T^2}{2K^2}$
• C. $\frac{2K}{T^2}$
• D. $\frac{2T^2}{K}$

Answer 1

The correct option is A

$\frac{T^2}{2K}$

Step 1: Given parameter

The spring extends by x on loading

Step 2: Parameter to be calculated

Energy stored by the spring

Step 3: Calculating the energy stored

The relation between the spring constant and tension is

$T=kx$

This implies that $x=\frac{T}{k}..........\left(1\right)$

Also, the energy stored in the spring is

$E=\frac{1}{2}k{x}^2$

Using equation (1), the expression of energy becomes

$E=\frac{1}{2}k\frac{T^2}{k^2}$

$E=\frac{T^2}{2k}$

Thus energy stored by the spring is $E=\frac{T^2}{2k}$

Hence, option (A) is correct.