Question:
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:
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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Springs in series carry the same force, so \(E = F^2/2k\) is inversely proportional to \(k\).
Updated On: Oct 31, 2026
- \(\frac{4}{3}\)
- \(\frac{16}{9}\)
- \(\frac{3}{4}\)
- \(\frac{9}{16}\)
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The Correct Option is A
Solution and Explanation
In series both springs carry the same force \(F\), so \(x = F/k\) and \(E = \frac12 kx^2 = \frac{F^2}{2k}\).
\(\frac{E_A}{E_B} = \frac{k_B}{k_A} = \frac{400}{300} = \frac43\).
(The total compression 8.75 cm is not needed for the ratio.)
Answer: (1) \(\frac43\)
\(\frac{E_A}{E_B} = \frac{k_B}{k_A} = \frac{400}{300} = \frac43\).
(The total compression 8.75 cm is not needed for the ratio.)
Answer: (1) \(\frac43\)
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