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Grade: 11

                        

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

2 years ago

Answers : (1)

Arun
24742 Points
							
Dear Balbir
 
Potential energy stored, when it is extended by x, is
U1= (1/2) k *x²
Potential energy stored, when it is further extended by y, is
U2 = (1/2) k* (x+y)²
Work done = U2 - U1
= (1/2) k [x² +y² + 2xy - x²]
= (1/2) k*y ( 2x +y)
 
Regards
Arun (askIITians forum expert)
2 years ago
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