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In a wire, when the elongation is 2 cm, the energy stored is E. if the wire is stretched by 10 cm, then the energy stored in the wire will be (a) E (b) 5E (c) 25E (d) 25/2E

Jayant Kumar , 12 Years ago
Grade 11
anser 1 Answers
Vikas TU

We know, potential energy of spring is given by (1/2)kx² where k is spring constant and x is stretched distance.

Here , x=2cm so, U=(1/2)k*2²=2k

For, x=10cm :-

E=(1/2)k*10²=50k=25(2k)=25U.

Final potential energy will be 25U.

Last Activity: 5 Years ago
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