A uniform rod of length l is acted upon by a force F in a...

The acceleration of the rod is

$a = \dfrac{F}{m}$

So, the tension in the rod at a distance $x$ from the free end is

$T\left(x\right) = \dfrac{mx}{l} \times \dfrac{F}{m} = \dfrac{Fx}{l}$

The longitudinal stress developed inside the rod at a distance $x$ from the free end is

$\sigma = \dfrac{T\left(x\right)}{A} = \dfrac{Fx}{lA}$

The strain energy density (energy per unit volume) is

$\dfrac{dU}{dV} = \dfrac{1}{2}\dfrac{\sigma2}{Y} = \dfrac{1}{2}\dfrac{F^{2}x^{2}}{YA^{2}l^{2}}$, where $dV = Adx$ is the volume of a small part of the rod

So the energy stored in the rod is

$U = \dfrac{1}{2}\dfrac{F2}{YA^{2}l^{2}}{∫}_{0}^{l}x^{2}dx$

$U = \dfrac{F^{2}l}{6A^{2}Y}$