Question

Two springs of spring constants $1500N{m}^{-1}$ and $3000N{m}^{-1}$ respectively are stretched with the same force. They will have potential energy in the ratio:

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is B

$2:1$

Step 1: Given

Spring constant for first spring: $k_1=1500N/m$

Spring constant for second spring: $k_2=3000N/m$

Force exerted on both springs: $F_1=F_2=F$

Step 2: Formula Used,

$F=kx$

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

Step 3: Find an expression for potential energy of first spring

Find an expression for displacement made by first spring using the formula

$$\begin{gathered} F_1=k_1{x}_1 \\ \Rightarrow F=1500x_1 \\ \Rightarrow x_1=\frac{F}{1500} \end{gathered}$$

Find an expression for potential energy of first spring using the formula and substituting the expression obtained for displacement.

$\begin{array}{rcl}{U}_1& =& \frac{1}{2}{k}_1{x_1}^2\\ & =& \frac{1}{2}\times 1500\times {\left(\frac{F}{1500}\right)}^2\\ & =& \frac{F^2}{3000}\end{array}$

Step 4: Find an expression for potential energy of second spring

Find an expression for displacement made by second spring using the formula

$$\begin{gathered} F_2=k_2{x}_2 \\ \Rightarrow F=3000x_2 \\ \Rightarrow x_2=\frac{F}{3000} \end{gathered}$$

Find an expression for potential energy of first spring using the formula and substituting the expression obtained for displacement.

$\begin{array}{rcl}{U}_2& =& \frac{1}{2}{k}_2{x_2}^2\\ & =& \frac{1}{2}\times 3000\times {\left(\frac{F}{3000}\right)}^2\\ & =& \frac{F^2}{6000}\end{array}$

Step 5: Calculate the ratio of potential energies by dividing the potential energy of first spring by the potential energy of second spring

$\begin{array}{rcl}\frac{U_1}{U_2}& =& \frac{\frac{F^2}{3000}}{\frac{F^2}{6000}}\\ & =& \frac{6000}{3000}\\ & =& \frac{2}{1}\end{array}$

Hence, the potential energies are in the ratio $2:1$.