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Chapter 6: Problem 50

(a) If the length of the Achilles tendon increases \(0.50 \mathrm{cm}\) when the force exerted on it by the muscle increases from \(3200 \mathrm{N}\) to $4800 \mathrm{N},$ what is the "spring constant" of the tendon? (b) How much work is done by the muscle in stretching the tendon \(0.50 \mathrm{cm}\) as the force increases from \(3200 \mathrm{N}\) to \(4800 \mathrm{N} ?\)

Short Answer

Expert verified
Answer: The spring constant of the Achilles tendon is 9600 N/m, and the work done by the muscle in stretching the tendon is 12 Joules.

Step by step solution

01

Understand Hooke's law

Hooke's law states that the force exerted on a spring (or any elastic object) is proportional to its elongation. Mathematically, this can be represented as: F = k * x, where F is the force exerted on the spring, k is the spring constant and x is the elongation of the spring.
02

Calculate the spring constant

As given in the problem, when the force increases from 3200 N to 4800 N, the elongation of the tendon is 0.50 cm. Using Hooke's law, we have: 3200 = k * 0 4800 = k * 0.50 Solving for k, we get: k = 4800 / 0.50 = 9600 N/m
03

Calculate the initial potential energy

The initial potential energy of the tendon can be calculated using the formula U = 1/2 * k * x^2. At the beginning, when the force is 3200 N, there is no elongation in the tendon (x = 0), so the initial potential energy is: U_initial = 1/2 * 9600 * 0^2 = 0 J
04

Calculate the final potential energy

When the force exerted on the tendon increases to 4800 N and the elongation is 0.50 cm (which we need to convert to meters, i.e., 0.005 m), we can calculate the final potential energy using the same formula: U_final = 1/2 * 9600 * 0.005^2 = 12 J
05

Calculate the work done by the muscle

Using the work-energy theorem, the work done by the muscle can be calculated as the difference between the final and initial potential energy: W = U_final - U_initial W = 12 J - 0 J = 12 J The work done by the muscle in stretching the tendon is 12 Joules.

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Most popular questions from this chapter

The orbit of Halley's comet around the Sun is a long thin ellipse. At its aphelion (point farthest from the Sun), the comet is $5.3 \times 10^{12} \mathrm{m}\( from the Sun and moves with a speed of \)10.0 \mathrm{km} / \mathrm{s} .$ What is the comet's speed at its perihelion (closest approach to the Sun) where its distance from the Sun is \(8.9 \times 10^{10} \mathrm{m} ?\)
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