Q11-24E - University Physics with Modern Physics

Question

A relaxed biceps muscle requires a force of 25.0 N for an elongation of ; the same muscle under maximum tension requires a force of 500 N - for the same elongation. Find Young’s modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0 cm².

Step-by-Step Solution

Answer

$$\begin{gathered} 3.33\times {10}^{4}Pa \\ 6.67\times {10}^{5}Pa \end{gathered}$$

Step 1: Given information:

$Force F = 25N, Elongation \Delta l=3.0cm, length l_0=0.2m and Area A=50c{m}^2$

Step 2: Concept/Formula used:

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

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

Step 3: Young modulus for relaxed condition:

$\begin{array}{rcl}A& =& 50c{m}^2=50\times {10}^{-4}{m}^2\\ Y& =& \frac{\left(0.2m\right)\left(25N\right)}{\left(50\times {10}^{-4}{m}^2\right)\left(3\times {10}^{-2}m\right)}\\ & =& 3.33\times {10}^{4}Pa\end{array}$

Step 4: Young modulus for maximum tension condition:

$\begin{array}{rcl}Y& =& \frac{\left(0.2m\right)\left(500N\right)}{\left(50\times {10}^{-4}{m}^2\right)\left(3\times {10}^{-2}m\right)}\\ & =& 6.67\times {10}^{5}Pa\end{array}$

Under maximum tension the muscle tissue is much more difficult to stretch.