Question

A $5kg$ rod of square cross section $5cm$ on a side and $1m$ long is pulled along a smooth horizontal surface by a force applied at one end. The rod has a constant acceleration of $2m/s^2$. Determine the elongation in the rod. (Young's modulus of the material of the rod is $5\times {10}^{3}N/m^9)$.

• A. Zero, as for elongation to be there, equal and opposite forces must act on the rod
• B. Non-zero but cannot be determine from the given situation
• C. $4\mu m$
• D. $16\mu m$

Answer 1

The correct option is A Zero, as for elongation to be there, equal and opposite forces must act on the rod

Let the force applied be $p$, then $P=ma$
$\Rightarrow p=5\times 2=10N$
For the element shown in the figure,
$T=\frac{m}{l}(l-x)\times a$
$\Rightarrow T=\frac{5}{l}(1-x)\times 2=10(1-x)$
Elongation in $dx$ is $\frac{△\left(dx\right)}{dx}=\frac{T/A}{A}$
$△\left(dx\right)=\frac{10(1-x)}{5\times {10}^{-2}{)}^2}\times \frac{1}{5\times {10}^9}dx$
Total elongation, $△l={\int }_0^l\frac{10(1-x)}{25\times {10}^{-4}}\times \frac{1}{5\times {10}^9}dx=0.4\times {10}^{-6}m$.