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A smooth uniform string of natural lengt...

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.

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(a) As gravity pulls the rod downward, there is tension developed in the. However, the lower points have lesser tension than the upper points, as they should support lesser weight.
Tension in the rod at distance x from the lower end is T, where
`T=((mx)/(l))g`.

Thus, stress at distance x from the lower end is `sigma =T//A` and strain at the point is
`S=(omega)/(Y)=(mxg//lA)/(Y)`.
We can write strain at distance x as ds//dx, here , ds is the change in length in an element of length dx.
`(ds)/(dx)=(mxg//lA)/(Y)`.
Therefore, the total change in the length of the rod,
`Delta L = int_(0)^(Delta L)ds = int_(0)^(i)((mgx)/(lAY))dx=(mg)/(lAY)((l^(2))/(2))`.
Hence, total strain in the rod is mg//2AY.
(b) Also find the elastic potential energy stored in the rod.
The energy stored in the element at distance x from the lower end is
`dU=(1)/(2)xx(mgx)/(lA)xx(ds)/(dx)xx Adx`.
As we know from Eq. 12-10,
`(ds)/(dx)=(mxg//lA)/(Y)`.
Therefore, the total elastic potential energy stored in the rod is
`U_(0)=int_(0)^(U_(0))dU=(m^(2)g^(2))/(2l^(2)AY)int_(0)^(l)x^(2)dx`
`=(m^(2)g^(2))/(2l^(2)AY)xx(l^(3))/(3)`
`=(m^(2)g^(2)l)/(6AY)`.
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