2001 AIPMT Work, Energy and PowerPotential Energy of a Spring Easy Question 7 of 7 in Potential Energy of a Spring

Two springs A and B having spring constants $K_A$ and $K_B$ ($K_A = 2K_B$) are stretched by applying force of equal magnitude. If energy stored in spring A is E then energy stored in B will be : -

A 2E
B $\dfrac{E}{4}$
C $\dfrac{E}{2}$
D 4E

The answer is A.

Solution

Energy stored in a spring: $E = \dfrac{1}{2}kx^2$. With force $F = kx$, $x = \dfrac{F}{k}$:
$E = \dfrac{kx^2}{2} = \dfrac{k^2x^2}{2k} = \dfrac{F^2}{2k}$
The force is equal, so $E \propto \dfrac{1}{k}$.
$\dfrac{E_B}{E_A} = \dfrac{K_A}{K_B} = 2 \Rightarrow E_B = 2E$

Concept Energy stored for equal force

Potential Energy of a Spring in past papers

7 questions from this chapter have appeared across 7 exam years.