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 , 10 Years ago
Grade 11
1 Answers
Vikas TU
Last Activity: 4 Years ago
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.
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