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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 its Young's modulus is Y, then the elastic potential energy stored in the rod due to elongation is



Correct Answer
A
U=F2l6AY
Your Answer
B
U=F2l3AY
Your Answer
C
U=F2l24A2Y
Your Answer
D
U=F2l22AY

Solution

The correct option is A U=F2l6AY

The acceleration of the rod is
a=Fm
So, the tension in the rod at a distance x from the free end is given by,
T(x)=mx×a
mx=λ×x, where λ=ml
T(x)=mxl×Fm=Fxl
The longitudinal stress developed inside the rod at a distance x from the free end is
σ=T(x)A=FxlA
The strain energy density (energy per unit volume) at distance x is
dUdV=12σ2Y=12F2x2YA2l2
[where dV=Adx is the volume of a small part of the rod]
dUAdx=12F2x2YA2l2

So the energy stored in the rod is found by integrating
dU=12F2YA2l2Ax2dx
U=12F2YAl2l0x2dx=F22AYl2×l33
U=F2l6AY

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