Question

A Plank of mass $M$ is suspended horizontally by using two wires as shown in the figure. They havesame length (L) and same cross-sectional area (A). Their Young's moduli are $Y_1$ and $Y_2$ respectively.The elastic potential energy of the system will be

• A. $\frac{2M^2{g}^{2}L}{A(Y_1+Y_2)}$
• B. $\frac{M^2{g}^{2}L}{A(Y_1+Y_2)}$
• C. $\frac{M^2{g}^{2}L}{2A(Y_1+Y_2)}$
• D. $\frac{M^2{g}^{2}L(Y_1+Y_2)}{2A{Y}_1{Y}_2}$

Answer 1

Answer: (C)

$\frac{M^2{g}^{2}L}{2A(Y_1+Y_2)}$

- $A=$ Areas of cross section

$L=$ length of wire

$Y_1{Y}_2=$ Young modulus of

$M=$mass of plank

$\begin{array}{c}\text{we define a quantity}\\ K=\frac{F}{\Delta L}=\frac{y_A}{L}\therefore (y=\frac{FL}{A\Delta L})\end{array}$

$\text{in case of parral}$

$\begin{array}{c}\text{Keq}=K_1+K_2\\ \frac{Y_{eq}\left(2A\right)}{L_1}=\frac{Y_{1}A}{L}+\frac{Y_{2}A}{L}\\ \text{2Yeq}=Y_1+Y_2\end{array}$

$Y_{eq}=\frac{Y_1+Y_2}{2}$

$\begin{array}{c}\text{Potential Energy}\left(U\right)=\frac{1}{2}\times \text{stress}x\text{strain}x\text{volume}\\ U=\frac{1}{2}\times \text{stress}\times \frac{\text{Stress}}{Y_{eq}}\times (2A\times L)\therefore Y=\frac{\text{Stress}}{\text{strain}}\end{array}$

$\begin{array}{c}=\frac{1}{2}{\left(\frac{F}{2A}\right)}^2\times \frac{1}{Yeq}\times (2A\cdot L)\\ U=\frac{M^2{g}^{2}L}{2A(Y_1+Y_2)}\end{array}$

$\begin{array}{c}\therefore F=mg\\ \therefore Y_{eq}=\frac{Y_1+Y_2}{2}\\ \therefore v=2AL\end{array}$