Solving Elasticity & Work Problems

[ppkjref]

Homework Statement

A wire of length L, Young's modulus Y, and cross sectional area A, is stretch elastically by an amount ∆L.

show that the work done in stretching a wire must be W = [YA(∆L)^2] / 2L

Homework Equations

∆L = (1/Y)(F/A)L
W = F * d (where displacement = ∆L)

The Attempt at a Solution

If ∆L = (1/Y)(F/A)L
F = (YA∆L)/L

Also if W = F * d (where displacement = ∆L)
W = F * ∆L
F = W/∆L

Then, (YA∆L)/L = W/∆L
[YA(∆L)^2] / L = W

However I do not see how W = [YA(∆L)^2] / 2L. I am missing the 2.

 

[PhanthomJay]

Homework Statement

A wire of length L, Young's modulus Y, and cross sectional area A, is stretch elastically by an amount ∆L.

show that the work done in stretching a wire must be W = [YA(∆L)^2] / 2L

Homework Equations

∆L = (1/Y)(F/A)L
W = F * d (where displacement = ∆L)

The Attempt at a Solution

If ∆L = (1/Y)(F/A)L
F = (YA∆L)/L

Also if W = F * d (where displacement = ∆L)
W = F * ∆L
F = W/∆L

Then, (YA∆L)/L = W/∆L
[YA(∆L)^2] / L = W

However I do not see how W = [YA(∆L)^2] / 2L. I am missing the 2.

That's because the applied force is not constant, it varies linearly from 0 at the start to a max value at (∆L), similar to a force in a spring which varies in the same manner (F=kx, Hooke's law). You can find the work done by using calculus and the definition of work done by a force, or since the force varies linearly with x, you can just take the average of the max and min force values.

 

[ppkjref]

Okay thanks for your help.