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In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm find the energy stored

Answer :

The answer is 25 E

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

Here , x=4 cm so, U=(1/2)k*4²=8k

For, x= 4cm :-

E=(1/2)k* 4²= 8k=U.

Final potential energy will be E= U.

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