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. The muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0cm^2.

Answer 1

Answer:

$3.3\times10^{4} Pa$

$6.67\times10^{5} Pa$

Explanation:

Complete statement of the question is

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 (Pa) for the muscle tissue under each of these conditions .The muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0 cm^2.

For relaxed muscle :

$F$ = force required = 25 N

$L$ = Normal length = 0.2 m = 20 cm

$\Delta L$ = elongation in length = 3 cm

$A$ = Area of cross-section = 50 cm² = 50 x 10⁻⁴ m²

$Y$ = Young's modulus for the muscle

Young's modulus is given as

$Y = \frac{FL}{A \Delta L} \\Y = \frac{(25)(20)}{(50\times10^{-4}) (3)} \\Y = 3.3\times10^{4} Pa$

Under maximum tension :

$F'$ = force required = 500 N

$L$ = Normal length = 0.2 m = 20 cm

$\Delta L$ = elongation in length = 3 cm

$A$ = Area of cross-section = 50 cm² = 50 x 10⁻⁴ m²

$Y$ = Young's modulus for the muscle

Young's modulus is given as

$Y = \frac{F'L}{A \Delta L} \\Y = \frac{(500)(20)}{(50\times10^{-4}) (3)} \\Y = 6.67\times10^{5} Pa$