Question

A relaxed biceps muscle requires a force of 25.4 N for an elongation of 3.20 cm; under maximum tension, the same muscle requires a force of 520 N for the same elongation. Find Young's modulus (Pa) for the muscle tissue under each of these conditions if the muscle can be modeled as a uniform cylinder with an initial length of 0.200 m and a cross-sectional area of 50.0 cm2.

Answer 1

Answer:

Y = 31750 Pa = 31.75 KPa (For 24.5 N force)

Y = 312500 Pa = 312.5 KPa (For 250 N force)

Explanation:

Since the elongation is constant. Therefore, the strain will remain the same in both cases:

$Strain = \frac{Elongation}{Original\ Length}\\\\Strain = \frac{0.032\ m}{0.2\ m}\\\\Strain = 0.16$

FOR A FORCE OF 25.4 N:

$Stress = \frac{Force}{Area}\\\\Stress = \frac{25.4\ N}{0.005\ m^2}\\\\Sress = 5080\ Pa = 5.08\ KPa$

Now, for Young's Modulus:

$Y = \frac{Stress}{Strain}\\\\Y = \frac{5080\ Pa}{0.16}$

Y = 31750 Pa = 31.75 KPa

FOR A FORCE OF 520 N:

$Stress = \frac{Force}{Area}\\\\Stress = \frac{250\ N}{0.005\ m^2}\\\\Sress = 50000\ Pa = 50\ KPa$

Now, for Young's Modulus:

$Y = \frac{Stress}{Strain}\\\\Y = \frac{50000\ Pa}{0.16}$

Y = 312500 Pa = 312.5 KPa