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JEE Main 2013 (9 Apr, Shift 1) · Spring & Elastic Energy

Spring & Elastic Energy question from JEE Main 2013 (9 Apr, Shift 1)

Q23JEE Main 2013 · 9 APR · SHIFT 1Medium

Spring & Elastic Energy

Two springs of force constants 300 N/m (Spring A) and 400 N/m (Spring B) are joined together in series. The combination is compressed by 8.75 cm. The ratio of energy stored in A and B is $\frac{E_A}{E_B}$ . Then $\frac{E_A}{E_B}$ is equal to:

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Answer: (a)

In series both springs carry the same force $F$ , so $E = \frac{F^2}{2k}$ .
$\frac{E_A}{E_B} = \frac{k_B}{k_A} = \frac{400}{300}$
So the answer is (a) $\frac{4}{3}$ .

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