Question

Write down the expression for the elastic potential energy of a stretched wire.

Answer 1

The work done in stretching the wire by dl,

dW = F.dl

The total work done in stretching the wire from 0 to l is

W = `int_0^"l" "F"."dl"` ..........(1)

From Young’s modulus of elasticity, force becomes,

Y = `"F"/"A" xx "L"/"l" = "YAl"/"L"` ........(2)

Substituting equation (2) in (1) we get,

W = `int_0^"l" "YAl"/"L" "dl"`

Since l is the dummy variable in the integration, we can change l to lʹ(not in limits).

Therefore W = `int_0^"l" "YAlʹ"/"L" "dlʹ" = "YA"/"L"["lʹ"^2/2]_0^"l" = "YA"/"L" "l"^2/2 = 1/2["YAl"/"L"] "l" = 1/2 "Fl"`

W = `1/2` Fl

This work done is known as the elastic potential energy of a stretched wire.