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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

A 75.0 -kg skier starts from rest and slides down a 32.0 -m frictionless slope that is inclined at an angle of \(15.0^{\circ}\) with the horizontal. Ignore air resistance. (a) Calculate the work done by gravity on the skier and the work done by the normal force on the skier. (b) If the slope is not frictionless so that the skier has a final velocity of \(10.0 \mathrm{m} / \mathrm{s},\) calculate the work done by gravity, the work done by the normal force, the work done by friction, the force of friction (assuming it is constant), and the coefficient of kinetic friction.
Sam pushes a \(10.0-\mathrm{kg}\) sack of bread flour on a frictionless horizontal surface with a constant horizontal force of \(2.0 \mathrm{N}\) starting from rest. (a) What is the kinetic energy of the sack after Sam has pushed it a distance of \(35 \mathrm{cm} ?\) (b) What is the speed of the sack after Sam has pushed it a distance of \(35 \mathrm{cm} ?\)
The power output of a cyclist moving at a constant speed of $6.0 \mathrm{m} / \mathrm{s}\( on a level road is \)120 \mathrm{W} .$ (a) What is the force exerted on the cyclist and the bicycle by the air? (b) By bending low over the handlebars, the cyclist reduces the air resistance to \(18 \mathrm{N} .\) If she maintains a power output of \(120 \mathrm{W},\) what will her speed be?
An airline executive decides to economize by reducing the amount of fuel required for long-distance flights. He orders the ground crew to remove the paint from the outer surface of each plane. The paint removed from a single plane has a mass of approximately \(100 \mathrm{kg} .\) (a) If the airplane cruises at an altitude of \(12000 \mathrm{m},\) how much energy is saved in not having to lift the paint to that altitude? (b) How much energy is saved by not having to move that amount of paint from rest to a cruising speed of $250 \mathrm{m} / \mathrm{s} ?$
Human feet and legs store elastic energy when walking or running. They are not nearly as efficient at doing so as kangaroo legs, but the effect is significant nonetheless. If not for the storage of elastic energy, a \(70-\mathrm{kg}\) man running at \(4 \mathrm{m} / \mathrm{s}\) would lose about \(100 \mathrm{J}\) of mechanical energy each time he sets down a foot. Some of this energy is stored as elastic energy in the Achilles tendon and in the arch of the foot; the elastic energy is then converted back into the kinetic and gravitational potential energy of the leg, reducing the expenditure of metabolic energy. If the maximum tension in the Achilles tendon when the foot is set down is \(4.7 \mathrm{kN}\) and the tendon's spring constant is $350 \mathrm{kN} / \mathrm{m},$ calculate how far the tendon stretches and how much elastic energy is stored in it.
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