Question

Two springs of spring constant $1000N{m}^{-1}$ and $2000N{m}^{-1}$ are stretched with same force. They will have potential energy in the ratio of:

• A. $2:1$
• B. $2^2:1^2$
• C. $1:2$
• D. $1^2:2^2$

Answer 1

The correct option is A $2:1$

$\text{Step 1: Elongation in spring}$

If a spring of spring constant $k$ is elongated by $x$ then spring force$\left(f\right)=kx$

$f=k_1{x}_1\Rightarrow x_1=\frac{f}{k_1}$

$f=k_2{x}_2\Rightarrow x_2=\frac{f}{k_2}$

$\text{Step 2: Potential energy of spring}$

If a spring of spring constant $k$ is elongated by $x$ then potential energy $\left(U\right)=\frac{1}{2}k{x}^2$

$U_1=\frac{1}{2}{k}_1{x}_1^2=\frac{1}{2}{k}_1\frac{f^2}{k_1^2}=\frac{f^2}{2k_1}$ $....\left(1\right)$

$U_2=\frac{1}{2}{k}_2{x}_2^2=\frac{f^2}{2k_2}$ $....\left(2\right)$

$\text{Step 3: Ratio of potential energy}$

Divide equation $\left(1\right)$ by equation $\left(2\right)$

Given $k_1=1000N/m,k_2=2000N/m$

$\frac{U_1}{U_2}=\frac{k_2}{k_1}=\frac{2000}{1000}=\frac{2}{1}$

Hence ratio of potential energy is $2:1$

Option $A$ is correct.