Question

A uniform rod of length $l$ is acted upon by a force $F$ in a gravity-free region, as shown in the figure. If the area of cross-section of the rod is $A$ and it's Young's modulus is $Y$, then the elastic potential energy stored in the rod due to elongation is

• A. $U=\frac{F^{2}I}{6Y}$
• B. $U=\frac{F^{2}l}{3Y}$
• C. $U=\frac{F^2{l}^2}{4Y}$
• D. $U=\frac{F^2{l}^2}{2Y}$

Answer 1

Answer: (A)

The acceleration of the rod is
$a=\frac{F}{m}$
So, the tension in the rod at a distance $x$ from the free end is
$T(x)=\frac{mx}{l}\times \frac{F}{m}=\frac{Fx}{l}$
The longitudinal stress developed inside the rod at a distance $x$ from the free end is
$\sigma =\frac{T(x)}{A}=\frac{Fx}{lA}$
The strain energy density (energy per unit volume) is
$\frac{dU}{dV}=\frac{1}{2}\frac{{\sigma }^2}{Y}=\frac{1}{2}\frac{F^2{x}^2}{Y{A}^2{l}^2}$, where $dV=Adx$ is the volume of a small part of the rod
So the energy stored in the rod is
$U=\frac{1}{2}{F}^2{\int }_0^1{x}^{2}dx$
$U=\frac{F^{2}l}{6Y}$