A smooth uniform, string of natural length `l`, cross-sectional area `A` and Young’s modulus `Y` is pulled along its length by a force `F` on a horizontal surface. Find the elastic potential energy stored in the string.

Answer

`U=(F^(2)l)/(6AY)`

Explanation

As discussed Mechanics the tension `T` in the string at a distance `x` from its free end is given
`T=F/lx`
Hence stress, `p=T/A=F/(Al)x`

Substituting `p` in the formula`U=1/(2Y)intp^(2)dV`
we have `U=1/(2Y)int_(0)^(l) (F^(2))/(A^(2)l^(2))x^(2)dV`, where d`V=Adx`
This gives `U=(F^(2)l)/(6AY)`