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Chapter 6: Problem 51

(a) If forces of magnitude \(5.0 \mathrm{N}\) applied to each end of a spring cause the spring to stretch \(3.5 \mathrm{cm}\) from its relaxed length, how far do forces of magnitude \(7.0 \mathrm{N}\) cause the same spring to stretch? (b) What is the spring constant of this spring? (c) How much work is done by the applied forces in stretching the spring \(3.5 \mathrm{cm}\) from its relaxed length?

Short Answer

Expert verified
a) The elongation caused by a 7.0 N force on the spring is approximately 4.9 cm. b) The spring constant of the spring is approximately 142.86 N/m. c) The work done by the applied forces in stretching the spring 3.5 cm from its relaxed length is approximately 0.088 J.

Step by step solution

01

Calculate the spring constant k with the given data

Using the given information, we can write Hooke's Law as: \(5.0 \mathrm{N} = k (3.5 \mathrm{cm})\) Convert the elongation from cm to m: \(3.5 \mathrm{cm} = 0.035 \mathrm{m}\) Then, we can solve for the spring constant k: \(k = \dfrac{5.0 \mathrm{N}}{0.035 \mathrm{m}} \approx 142.86 \mathrm{N/m}\)
02

Calculate the elongation caused by a 7.0 N force

Using the calculated value of k in Hooke's Law: \(7.0 \mathrm{N} = 142.86 \mathrm{N/m} \cdot x\) Now, we can solve for the elongation x: \(x = \dfrac{7.0 \mathrm{N}}{142.86 \mathrm{N/m}} \approx 0.049 \mathrm{m}\) Converting the elongation from m to cm: \(x \approx 4.9 \mathrm{cm}\) Therefore, a force of 7.0 N causes the spring to stretch 4.9 cm from its relaxed length.
03

Calculate the work done by the applied forces in stretching the spring 3.5 cm from its relaxed length

We can use the formula for work done on a spring: \(W = \dfrac{1}{2}kx^2\) Substituting the values of k and x: \(W = \dfrac{1}{2}(142.86 \mathrm{N/m})(0.035 \mathrm{m})^2\) Now, we can calculate the work done: \(W \approx 0.088 \mathrm{J}\) Therefore, the work done by the applied forces in stretching the spring 3.5 cm from its relaxed length is approximately 0.088 J.

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Most popular questions from this chapter

You shoot a \(51-\mathrm{g}\) pebble straight up with a catapult whose spring constant is \(320 \mathrm{N} / \mathrm{m} .\) The catapult is initially stretched by \(0.20 \mathrm{m}\). How high above the starting point does the pebble fly? Ignore air resistance.
How many grams of carbohydrate does a person of mass 74 kg need to metabolize to climb five flights of stairs \((15 \mathrm{m}\) height increase)? Each gram of carbohydrate provides 17.6 kJ of energy. Assume \(10.0 \%\) efficiency-that is, \(10.0 \%\) of the available chemical energy in the carbohydrate is converted to mechanical energy. What happens to the other \(90 \%\) of the energy?
Hilda holds a gardening book of weight \(10 \mathrm{N}\) at a height of $1.0 \mathrm{m}\( above her patio for \)50 \mathrm{s}$. How much work does she do on the book during that 50 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.
Two springs with spring constants \(k_{1}\) and \(k_{2}\) are connected in series. (a) What is the effective spring constant of the combination? (b) If a hanging object attached to the combination is displaced by \(4.0 \mathrm{cm}\) from the relaxed position, what is the potential energy stored in the spring for \(k_{1}=5.0 \mathrm{N} / \mathrm{cm}\) and $k_{2}=3.0 \mathrm{N} / \mathrm{cm} ?$ [See Problem \(83(\mathrm{a}) .]\)
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