Questions: The length of the Achilles tendon increases 0.500 cm when the force exerted on it by the muscle increases from 3214 N to 4784 N. How much work is done by the muscle in stretching the tendon?

Solution Steps

We are given:

• The initial force \( F_1 = 3214 \, \text{N} \)
• The final force \( F_2 = 4784 \, \text{N} \)
• The increase in length of the tendon \( \Delta x = 0.500 \, \text{cm} = 0.005 \, \text{m} \)

The work done by the muscle can be calculated using the average force exerted over the distance the tendon is stretched. The average force \( F_{\text{avg}} \) is given by: \[ F_{\text{avg}} = \frac{F_1 + F_2}{2} \] Substituting the given values: \[ F_{\text{avg}} = \frac{3214 \, \text{N} + 4784 \, \text{N}}{2} = \frac{7998 \, \text{N}}{2} = 3999 \, \text{N} \]

The work done \( W \) by the muscle in stretching the tendon is given by: \[ W = F_{\text{avg}} \cdot \Delta x \] Substituting the values: \[ W = 3999 \, \text{N} \cdot 0.005 \, \text{m} = 19.995 \, \text{J} \]

Final Answer