A uniform elastic plank moves due to a constant force `F_(0)` applied at one end whose area is `S`. The Young's modulus of the plank is `Y`. The strain produced in the direction of force is

Explanation

Let us consider an element of the rod at a distance x from the free end (figure). For the considered element `'T-T'` are internal restoring forces which produce elongation and `dT` provides the acceleration to the element. For the element from Newton's law:
`dT=(dm)w=(m/ldx)F_o/m=F_o/ldx`
As free end has zero tension, on integrating the above expression,
`underset(o)overset(T)intdT=F_0/lunderset(o)overset(x)intdx` or `T=F_o/lx`
Elongation in the considered element of length `dx` :
`deltaxi=(sigma)/(E)(x)dx=(T)/(SE)dx=(F_(xo)xdx)/(SEL)`
Thus total elongation `xi=(F_o)/(SEl)underset(x)overset(l)intdx=(F_ol)/(2SE)`
Hence the sought strain
`sigma=(xi)/(l)=(F_o)/(2SE)`