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Chapter 6: Problem 29

A force of 160 \(\mathrm{N}\) stretches a spring 0.050 \(\mathrm{m}\) beyond its unstretched length. (a) What magnitude of force is required to stretch the spring 0.015 \(\mathrm{m}\) beyond its unstretched length? To compress the spring 0.020 \(\mathrm{m} ?\) (b) How much work must be done to stretch the spring 0.015 \(\mathrm{m}\) beyond its unstretched length? To compress the spring 0.020 \(\mathrm{m}\) from its unstretched length?

Short Answer

Expert verified
(a) 48 N to stretch 0.015 m, 64 N to compress 0.020 m. (b) 0.36 J to stretch, 0.64 J to compress.

Step by step solution

01

Understanding Hooke's Law

Hooke's Law states that the force \( F \) required to stretch or compress a spring by a distance \( x \) is directly proportional to that distance: \( F = kx \), where \( k \) is the spring constant.
02

Find the Spring Constant

We know that a force of 160 \( N \) stretches the spring by 0.050 \( m \). Using Hooke's Law, we solve for \( k \): \( 160 = k \, \times \, 0.050 \). Therefore, \( k = \frac{160}{0.050} = 3200 \, N/m \).
03

Calculate the Force to Stretch 0.015 m

Using the spring constant \( k = 3200 \, N/m \), calculate the required force to stretch the spring 0.015 \( m \): \( F = 3200 \, \times \, 0.015 = 48 \, N \).
04

Calculate the Force to Compress 0.020 m

To find the force to compress by 0.020 \( m \), we again use Hooke's Law: \( F = kx = 3200 \, \times \, 0.020 = 64 \, N \). The magnitude is the same, but the direction is opposite (compression instead of extension).
05

Work Done in Stretching 0.015 m

The work done \( W \) in stretching or compressing a spring is given by \( W = \frac{1}{2}kx^2 \). Substituting the values gives: \( W = \frac{1}{2} \, \times \, 3200 \, \times \, (0.015)^2 = 0.36 \, J \).
06

Work Done in Compressing 0.020 m

Use the same formula for work: \( W = \frac{1}{2} \, \times \, 3200 \, \times \, (0.020)^2 = 0.64 \, J \). The work formula is the same for both stretching and compressing.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Spring Constant
The spring constant, represented as \(k\), is a crucial part of understanding how springs behave under force. It essentially measures the stiffness of a spring. The higher the spring constant, the stiffer the spring is. To find \(k\), we rely on experiments that determine how much force needs to be applied to stretch or compress the spring by a certain amount. In the given exercise, a force of 160 \(N\) stretches the spring by 0.050 \(m\). Applying Hooke鈥檚 Law, expressed as \(F = kx\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the distance the spring is stretched or compressed, we can solve for the spring constant:
  • Given, \(F = 160\, N\) and \(x = 0.050\,m\).
  • Spring constant, \(k = \frac{160}{0.050} = 3200\,N/m\).
This value tells us how much force in Newtons is needed to stretch or compress the spring by one meter.
Calculating Force on the Spring
For any spring, calculating the force needed to either stretch or compress it involves using the spring constant \(k\) and the desired distance \(x\) referenced in Hooke's Law. For instance, if we want to stretch the spring by 0.015 meters, we calculate the force as follows:
  • \(F = kx = 3200 \, \times \, 0.015 = 48\, N\).
Similarly, if the spring is to be compressed by 0.020 meters, the calculation becomes:
  • \(F = kx = 3200 \, \times \, 0.020 = 64\, N\).
It's important to note that while the magnitude of the force remains the same for both stretching and compressing, the direction is considered opposite.
Work Done on the Spring
The concept of work done is linked to the energy transferred to or from the spring when it is stretched or compressed. Work done, \(W\), can be calculated using the formula \(W = \frac{1}{2} k x^2\). This formula considers the spring's stiffness and the distance moved.For stretching by 0.015 meters:
  • \(W = \frac{1}{2} \times 3200 \times (0.015)^2 = 0.36\, J\).
The same formula applies to compressing the spring. For a 0.020-meter compression:
  • \(W = \frac{1}{2} \times 3200 \times (0.020)^2 = 0.64\, J\).
Whether stretching or compressing, this work transfers energy into or out of the spring, with no difference in the calculation except for the displacement involved.

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Most popular questions from this chapter

Meteor Crater. About \(50,000\) years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Recent \((2005)\) measurements estimate that this meteor had a mass of about \(1.4 \times 10^{3} \mathrm{kg}\) (around \(150,000\) tons) and hit the ground at 12 \(\mathrm{km} / \mathrm{s}\). (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0 -megaton nuclear bomb? (A megaton bomb releases the same energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9} \mathrm{J}\) of energy.)

Half of a Spring. (a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant \(k\) , what is the force constant of each half, in terms of \(k ?\) (Hint: Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal? (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k ?\)

An old oaken bucket of mass 6.75 \(\mathrm{kg}\) hangs in a well at the end of a rope. The rope passes over a frictionless pulley at the top of the well, and you pull horizontally on the end of the rope to raise the bucket slowly a distance of 4.00 \(\mathrm{m} .\) (a) How much work do you do on the bucket in pulling it up? (b) How much work does gravity do on the bucket? (c) What is the total work done on the bucket?

Rotating Bar. A thin, uniform \(12.0-\mathrm{kg}\) bar that is 2.00 \(\mathrm{m}\) long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (Hint: Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass \(d m\) and integrate to add up the kinetic energy of all these segments.)

A block of ice with mass 6.00 \(\mathrm{kg}\) is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force \(\vec{F}\) to it. As a result, the block moves along the \(x\) -axis such that its position as a function of time is given by \(x(t)=\alpha t^{2}+\beta t^{3}\) , where \(\alpha=0.200 \mathrm{m} / \mathrm{s}^{2}\) and \(\beta=0.0200 \mathrm{m} / \mathrm{s}^{3}\) . (a) Calculate the velocity of the object when \(t=4.00 \mathrm{s}\) . (b) Calculate the magnitude of \(\overrightarrow{\boldsymbol{F}}\) when \(t=4.00 \mathrm{s}\) . (c) Calculate the work done by the force \(\overrightarrow{\boldsymbol{F}}\) during the first 4.00 \(\mathrm{s}\) of the motion.
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