Q11-27E - University Physics with Modern Physics

Question

A metal rod that is 4.00 m long and 0.50 cm² 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

$Length l_0=4.00m, Area A=0.50c{m}^2, Elongation ∆I=0.20cm, Tension Force f=4000N$

Step 2: Concept/Formula used

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

Where, Y is Young’s modulus, I₀ is length of muscle, F is muscle force, A is cross-sectional area and $\mathbf{∆}\mathit{l}$ is elongation.

Step 3: Young’s modulus Calculation

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