1.

An elastic spring of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is

Answer»

`1/2 KY^2`
`1/2 k(x^2 + y^2)`
`1/2 k(x + y)^2`
`1/2 ky (2x + y)`

Solution :POTENTIAL energy stored in the spring when it is further extended by y is `U_2 = 1/2 k(x + y)^2`
`:.`WORK done = `U_2 - U_1 = 1/2 k(x + y)^2 - 1/2 kx^2`
`= 1/2 ky (2x + y)`.


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