Two springs A and B (kA =2kB ) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, then in B is (assume equilibrium):
Correct answer: B. 2E
- A. E/2
- B. 2E
- C. E
- D. E/4
Explanation
EA = ½ kAxA2 Keeping force constant and using F= kx As kA = 2kB Then 2xA=xB Energy stored in B will be: EB= ½ (kA/2) (2xA)2 EB= ½ (2xA2 kA) Replace EA = ½ kAxA2 EB = 2 EA Hence only B is correct (A, C, D give wrong answer)
This question appeared on the ETEA MDCAT 2019 paper, which you can sit online with every answer explained.
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Potential energy is energy stored because of position or configuration, including gravitational potential energy near Earth's surface and elastic energy in a stretched or compressed spring. Questions use changes in potential energy and conservation of mechanical energy, while distinguishing a chosen reference level from an absolute zero of energy.
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