Q24 E - University Physics with Modern Physics

Question

A relaxed biceps muscle requires a force of 25.0 N for an elongation of 3.0 cm; 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 $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{2}}$.

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 = 25 N, 

Elongation: $∆l=30.0 cm$,

length: $l_0=0.2 m$ and

Area: $A=50 c{m}^2$ .

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: (a) Young modulus for relaxed condition:

 $$\begin{gathered} A=50 c{m}^2 \\ =50\times {10}^{-4} m^2 \end{gathered}$$

$$\begin{gathered} Y=\frac{(0.2 m)(25 N)}{(50\times {10}^{-4} m^2)(3\times {10}^{-2} m)} \\ =3.33\times {10}^4 Pa \end{gathered}$$ 

Step 4: (b) Young modulus for maximum tension condition:

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

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