Question

Prove that work done in a long steel wire is equal to the half of the product of load and elongation

Answer 1

∫dW = limits of integration 0 to l ∫(Y*A*x)/L dx

W = [(Y*A)/L]*(limits of integration 0 to l)∫ x dx

W = [(Y*A)/L] *[x²/2] limits of integration 0 to l

W = [(Y*A)/L]* [l²/2]

W = [(Y*A*l)/L]* [l/2]      ↔(1)

Since we've got a relation ; F = (Y*A*x)/L

where wire is stretched fully @ x=l and it is known that F=mg

so the relation modifies ⇒ F = mg= (Y*A*x)/L = (Y*A*l)/L

⇒ mg = (Y*A*l)/L

putting the above relation in equation (1)

W = mg* (l/2)

W = 1/2*mg *l = 1/2 F*l

W = 1/2 *F*l

Hence proved