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An elastic spring of unstretched length l and force constant K is stretched by amount x it is further stretched by another length y the work done in the second stretching is

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3 years ago

## Answers : (1)

Arun
25768 Points
```							Dear Balbir Potential energy stored, when it is extended by x, isU1= (1/2) k *x²Potential energy stored, when it is further extended by y, isU2 = (1/2) k* (x+y)²Work done = U2 - U1= (1/2) k [x² +y² + 2xy - x²]= (1/2) k*y ( 2x +y) RegardsArun (askIITians forum expert)
```
3 years ago
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