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Chapter 6: Problem 100

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 \(83(\mathrm{a}) .]\)

Short Answer

Expert verified
Answer: The potential energy stored in the springs is \(15 \, J\).

Step by step solution

01

Determine the effective spring constant of springs in series

Consider two springs with spring constants \(k_1\) and \(k_2\) connected in series. The effective spring constant \(K\) can be calculated using the formula: \[ \frac{1}{K} = \frac{1}{k_1} + \frac{1}{k_2} \] Rearrange the equation to solve for \(K\): \[ K = \frac{k_1k_2}{k_1 + k_2} \] We will use this equation to find the effective spring constant of the combination.
02

Calculate the effective spring constant for the given values

Now, we will substitute the given values \(k_1 = 5.0 \, N/cm\) and \(k_2 = 3.0 \, N/cm\) into the equation: \[ K = \frac{(5.0)(3.0)}{5.0 + 3.0} = \frac{15}{8} \, N/cm \] The effective spring constant of the springs in series is \(\frac{15}{8} \, N/cm\).
03

Calculate the potential energy stored in the spring for the given displacement

The potential energy \(U\) stored in a spring can be calculated using the formula: \[ U = \frac{1}{2} K x^{2} \] where \(K\) is the effective spring constant and \(x\) is the displacement from the relaxed position. We will use the effective spring constant found in step 2 and the given displacement of \(4.0 \, cm\) to calculate the potential energy stored in the spring.
04

Calculate the potential energy stored in the spring for the given displacement and effective spring constant

Now, substitute the effective spring constant \(K = \frac{15}{8} \, N/cm\) and the displacement \(x = 4.0 \, cm\) into the formula for the potential energy: \[ U = \frac{1}{2} \times \frac{15}{8} \times (4.0)^{2} = \frac{15}{8} \times 8 = 15 \, J \] The potential energy stored in the spring is \(15 \, J\).

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Most popular questions from this chapter

The escape speed from the surface of the Earth is $11.2 \mathrm{km} / \mathrm{s} .$ What would be the escape speed from another planet of the same density (mass per unit volume) as Earth but with a radius twice that of Earth?
The escape speed from the surface of Planet Zoroaster is $12.0 \mathrm{km} / \mathrm{s} .$ The planet has no atmosphere. A meteor far away from the planet moves at speed \(5.0 \mathrm{km} / \mathrm{s}\) on a collision course with Zoroaster. How fast is the meteor going when it hits the surface of the planet?
The length of a spring increases by \(7.2 \mathrm{cm}\) from its relaxed length when a mass of \(1.4 \mathrm{kg}\) is hanging in equilibrium from the spring. (a) What is the spring constant? (b) How much elastic potential energy is stored in the spring? (c) A different mass is suspended and the spring length increases by \(12.2 \mathrm{cm}\) from its relaxed length to its new equilibrium position. What is the second mass?
The bungee jumper of Example 6.4 made a jump into the Gorge du Verdon in southern France from a platform \(182 \mathrm{m}\) above the bottom of the gorge. The jumper weighed \(780 \mathrm{N}\) and came within \(68 \mathrm{m}\) of the bottom of the gorge. The cord's unstretched length is \(30.0 \mathrm{m}\) (a) Assuming that the bungee cord follows Hooke's law when it stretches, find its spring constant. [Hint: The cord does not begin to stretch until the jumper has fallen \(30.0 \mathrm{m} .]\) (b) At what speed is the jumper falling when he reaches a height of \(92 \mathrm{m}\) above the bottom of the gorge?
Jane is running from the ivory hunters in the jungle. Cheetah throws a 7.0 -m-long vine toward her. Jane leaps onto the vine with a speed of $4.0 \mathrm{m} / \mathrm{s} .$ When she catches the vine, it makes an angle of \(20^{\circ}\) with respect to the vertical. (a) When Jane is at her lowest point, she has moved downward a distance \(h\) from the height where she originally caught the vine. Show that \(h\) is given by \(h=L-L \cos 20^{\circ},\) where \(L\) is the length of the vine. (b) How fast is Jane moving when she is at the lowest point in her swing? (c) How high can Jane swing above the lowest point in her swing?
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