Question

A smooth uniform string of natural length l, cross – sectional area A and Young’s modulus Y is pulled along its length by a force F on a horizontal surface. If the elastic potential energy stored in the string is $U={\displaystyle \frac{F^{2}l}{xAY}}$, Find x.

Answer 1

Hint: According to Mohr’s Law,
$Y={\displaystyle \frac{Stress}{Strain}}$
Elastic Potential Energy is given by the formula,
$U=\int dW$

Complete step by step solution:
It is given in the question that,
Natural length of the string is l
Cross – sectional area of the string is A
Young’s Modulus of the string is Y
Force acting on the string is F

According to Mohr’s Law,
$Y={\displaystyle \frac{Stress}{Strain}}$ ……….(1)
Where,
Y is the Young’s Modulus
Also,
$Stress={\displaystyle \frac{F}{A}}$
Where,
F is the Tension in the String
And,
$$Strain={\displaystyle \frac{\mathrm{\Delta }l}{l}}$$
Inserting the values of stress and strain in equation 1,
We get,
$=>Y={\displaystyle \frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{\mathrm{\Delta }l}{l}}}}$
$=>Y={\displaystyle \frac{Fl}{A\mathrm{\Delta }l}}$
$=>F={\displaystyle \frac{YA\mathrm{\Delta }l}{l}}$ ……..(2)
Elastic Potential Energy is given by the formula,
$U=\int dW$
Where dW is the work done
In the above formula,
$dW=F.dx$
Where,
dx is the change in length of the string due to the applied tension
Inserting the value of dW in above equation,
We get,
$=>U=\int F.dx$
Inserting the value of F in the above equation and the limits would be from 0 to $\mathrm{\Delta }l$ (Since, initially the string was unstretched so initial limit is 0 and finally the string is stretch by $\mathrm{\Delta }l$ so the final limit is $\mathrm{\Delta }l$)

We get,
$=>U=\underset{0}{\overset{\mathrm{\Delta }l}{\int }}{\displaystyle \frac{YAx}{l}}.dx$
$=>U={\displaystyle \frac{YA}{l}}\underset{0}{\overset{\mathrm{\Delta }l}{\int }}x.dx$
$=>U={\displaystyle \frac{YA}{l}}[{\displaystyle \frac{x^2}{2}}{]}_0^{\mathrm{\Delta }l}$
$=>U={\displaystyle \frac{YA}{l}}{\displaystyle \frac{{(\mathrm{\Delta }l)}^2}{2}}$ …………(3)
From Equation 2,
$=>F={\displaystyle \frac{YA\mathrm{\Delta }l}{l}}$
$=>\mathrm{\Delta }l={\displaystyle \frac{Fl}{AY}}$
Inserting the value of $\mathrm{\Delta }l$ in Equation 3,
We get,
$=>U={\displaystyle \frac{YA}{l}}{\displaystyle \frac{{({\displaystyle \frac{Fl}{AY}})}^2}{2}}$
$=>U={\displaystyle \frac{F^{2}l}{2AY}}$
Comparing the above equation with the one given in the question,
$=>U={\displaystyle \frac{F^{2}l}{2AY}}={\displaystyle \frac{F^{2}l}{xAY}}$
Clearly, we can see from the above expression that,
$x=2$

Note: Such types of questions are categorized into tricky sections. One requires in-depth knowledge of mechanics and calculus in order to solve such Calculation Intensive Problems.