A wire is stretched by a force 'F' which causes an extension ∆l, the energy stored in the wire is:
Correct answer: D. ½ F∆l
- A. F∆l
- B. 2F∆l
- C. ½ F∆l2
- D. ½ F∆l
Explanation
Given that, force applied = F . extension = ∆l and we assume the length of the wire to be = L.If the elastic limit is not exceeded then the stress is directly proportional to strain.Where stress is the amount of force applied per unit area (σ = F/A)And strain is extension per unit length (ε = ∆l/l)Hence,Energy stored = ½ x stress x strain x volumeEnergy stored = ½ x (F/A) x (∆l / L) x A x LEnergy stored = ½ F∆l
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