A wire has a cross sectional area ‘A’, length ‘L’ and young’s modulus ‘Y’. It is pulled by a force ‘F’ which causes a total extension of length ‘l’. The force F is so adjusted that the wire is only slowly stretched. Find the work done by the force in pulling the string by a length ‘dx’ when extension is x (0 > x > l).

A wire has a cross sectional area ‘A’, length ‘L’ and young’s modulus ‘Y’. It is pulled by a force ‘F’ which causes a total extension of length ‘l’. The force F is so adjusted that the wire is only slowly stretched. Find the work done by the force in pulling the string by a length ‘dx’ when extension is x (0 > x > l). Correct Answer (AY/L)xdx

The force F at any instant will be equal to internal force developed as the wire is to be slowly stretched. ∴ F=(AY/L)x ∴ Work done for dx extension = F*dx = (AY/L)xdx.

Related Questions

The change in length due to a tensile or compressive force acting on a body is given by (where P = Tensile or compressive force acting on the body, $$l$$ = Original length of the body, A = Cross-sectional area of the body and E = Young's modulus for the material of the body)