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Chapter 6: Problem 53

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?

Short Answer

Expert verified
Answer: 0.4959 J

Step by step solution

01

(a) Determine the force applied on the spring by the hanging mass

To calculate the constant spring, we first need to find the force applied on the spring by the mass hanging in equilibrium. Using the formula \(F = mg\), where \(F\) is the force, \(m\) is the mass, and \(g\) is the acceleration due to gravity (\(9.81 m/s^2\)), we can find the force applied: \(F = 1.4 * 9.81 = 13.734 N\)
02

(a) Calculate the spring constant using Hooke's law

Hooke's law states that the force applied on a spring is proportional to the displacement of the spring, or \(F = k\Delta{L}\), where \(F\) is the force, \(k\) is the spring constant, and \(\Delta{L}\) is the change in length of the spring. Rearranging the formula for \(k\), we get \(k = \frac{F}{\Delta{L}}\). Plug in the values: \(k = \frac{13.734 N}{0.072 m} = 190.75 N/m\)
03

(b) Calculate the elastic potential energy stored in the spring

Elastic potential energy stored in a spring is given by the formula \(PE = \frac{1}{2} k (\Delta{L})^2\). Using the spring constant obtained in part (a) and the change in length, we can find the elastic potential energy: \(PE = \frac{1}{2} (190.75 N/m) (0.072 m)^2 = 0.4959 J\)
04

(c) Determine the force applied on the spring by the second mass with the new spring length

The new equilibrium position increased the spring length by \(0.122 m\), we can use Hooke's law \(F = k\Delta{L}\) again to determine the force: \(F = (190.75 N/m) * (0.122 m) = 23.2515 N\)
05

(c) Calculate the second mass using the force obtained from the new spring length

Now that we have the force applied by the second mass, we can use the equation \(F = mg\) to find the mass. Rearrange the equation for \(m\): \(m = \frac{F}{g} = \frac{23.2515 N}{9.81 m/s^2} = 2.369 kg\) The spring constant is \(190.75 N/m\), the elastic potential energy stored in the spring is \(0.4959 J\), and the second mass is \(2.369 kg\).

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Most popular questions from this chapter

An ideal spring has a spring constant \(k=20.0 \mathrm{N} / \mathrm{m}\) What is the amount of work that must be done to stretch the spring \(0.40 \mathrm{m}\) from its relaxed length?
A car moving at \(30 \mathrm{mi} / \mathrm{h}\) is stopped by jamming on the brakes and locking the wheels. The car skids 50 ft before coming to rest. How far would the car skid if it were initially moving at $60 \mathrm{mi} / \mathrm{h} ?$ [Hint: You will not have to do any unit conversions if you set up the problem as a proportion. \(]\)
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A 4.0 -kg block is released from rest at the top of a frictionless plane of length \(8.0 \mathrm{m}\) that is inclined at an angle of \(15^{\circ}\) to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point \(5.0 \mathrm{m}\) along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer? A 4.0 -kg block is released from rest at the top of a frictionless plane of length \(8.0 \mathrm{m}\) that is inclined at an angle of \(15^{\circ}\) to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point \(5.0 \mathrm{m}\) along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer?
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