Q27 E - University Physics with Modern Physics

Question

A metal rod that is 4.00 m long and $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{2}}$ in cross-sectional area is found to stretch 0.20 cm under a tension of 5000 N. What is Young’s modulus for this metal?

Step-by-Step Solution

Answer

$2\times {10}^{11} Pa$

Step 1: Given information

Length: $l_0=4.00 m$,

Area: $A=0.50 c{m}^2$,

Elongation: $∆l=0.20 cm$,

Tension Force: F = 4000 N .

Step 2: Concept/Formula used

$\mathit{Y}\mathbf{=}\frac{{\mathbf{l}}_{\mathbf{0}}\mathbf{F}}{\mathbf{∆}\mathbf{l}}$

Where, Y is Young’s modulus, ${\mathbf{l}}_{\mathbf{0}}$ is length of muscle,  F is muscle force, A is cross-sectional area and $\mathbf{∆}\mathbf{l}$ is elongation.

Step 3: Young’s modulus Calculation

$$\begin{gathered} Y=\frac{(4.00 m)(5000 N)}{(0.50\times {10}^{-4} m^2)(0.20\times {10}^{-2} m)} \\ =2\times {10}^{11} Pa \end{gathered}$$