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Chapter 6: Problem 36

A spring of force constant \(300.0 \mathrm{~N} / \mathrm{m}\) and unstretched length \(0.240 \mathrm{~m}\) is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to \(15.0 \mathrm{~N}\). How long will the spring now be, and how much work was required to stretch it that distance?

Short Answer

Expert verified
The length the spring will now be is 0.290 m, and the work required to stretch it that distance is 0.375 J.

Step by step solution

01

Calculating the Stretched Length

Using Hooke's law, which can be written as \( F = kx \), where 'F' is the force, 'k' is the spring constant and 'x' is the displacement from the spring's natural length, the displacement can be found by rearranging the formula to \( x = F/k \). Substituting the given force \( F = 15.0 \, \mathrm{N} \) and spring constant \( k = 300.0 \, \mathrm{N/m} \), the displacement \( x = 15.0 \, \mathrm{N} / 300.0 \, \mathrm{N/m} = 0.05 \, \mathrm{m} \). The total length the spring now will be is the unstretched length plus the displacement, which is \( 0.240 \, \mathrm{m} + 0.05 \, \mathrm{m} = 0.290 \, \mathrm{m} \).
02

Calculating the Work Done

The work done in stretching the spring is given by the formula \( W = 1/2 * k * x^2 \), where 'W' is the work done, 'k' is the spring constant and 'x' is the displacement from the spring's natural length. Substituting the given spring constant \( k = 300.0 \, \mathrm{N/m} \) and calculated displacement \( x = 0.05 \, \mathrm{m} \), the work done \( W = 1/2 * 300.0 \, \mathrm{N/m} * (0.05 \, \mathrm{m})^2 = 0.375 \, \mathrm{J} \).

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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, often denoted as "k," plays a crucial role in determining how much a spring stretches or compresses under a given force. It represents the stiffness of the spring:
  • A high spring constant means the spring is stiff and requires more force to stretch.
  • A low spring constant indicates the spring is easily stretched.
In Hooke's Law, the spring constant units are expressed as Newtons per meter (N/m). This unit measures how much force is needed to stretch the spring by one meter. For our problem, the spring constant is 300.0 N/m. This signifies that for every meter stretch, 300.0 Newtons of force is required. Understanding this value helps in predicting the spring's behavior under force.
Exploring Work Done on the Spring
Work done is a concept integral to stretching or compressing a spring. It refers to the energy required to do so. The mathematical expression for work done, when dealing with springs, utilizes the formula:\[ W = \frac{1}{2} k x^2 \]Where:
  • "W" is the work done.
  • "k" is the spring constant.
  • "x" is the displacement of the spring from its natural length.
For the problem at hand, after substituting the values, the work done to stretch the spring by 0.05 meters is calculated to be 0.375 Joules. This calculation shows how much energy is required to overcome the spring's tension. It's important because it measures the effort needed to change the spring's state.
The Role of Displacement in Hooke's Law
Displacement in the context of springs refers to how much a spring is stretched or compressed from its normal length. It is a crucial variable in Hooke's Law, represented by "x" in the formula:\[ F = kx \]Where:
  • "F" is the force applied.
  • "k" is the spring constant.
  • "x" is the displacement of the spring.
In practical terms, displacement is the change in the spring's length when forces are applied. In our example, the displacement was calculated as 0.05 meters after applying a force of 15.0 N with a spring constant of 300.0 N/m. This means the spring's length increased by 0.05 meters from its natural length. Understanding displacement helps estimate how much a spring will stretch under various forces.

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Most popular questions from this chapter

An elevator has mass \(600 \mathrm{~kg},\) not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of \(20.0 \mathrm{~m}\) (five floors) in \(16.0 \mathrm{~s}\), and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass \(65.0 \mathrm{~kg}\).

Meteor Crater. About 50,000 years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Measurements from 2005 estimate that this meteor had a mass of about \(1.4 \times 10^{8} \mathrm{~kg}\) (around 150,000 tons) and hit the ground at a speed of \(12 \mathrm{~km} / \mathrm{s}\). (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 amount of energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9} \mathrm{~J}\) of energy.)

You are a member of an Alpine Rescue Team. You must project a box of supplies up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathrm{k}}\). Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{k},\) and \(\alpha\)

One end of a horizontal spring with force constant \(76.0 \mathrm{~N} / \mathrm{m}\) is attached to a vertical post. A \(2.00 \mathrm{~kg}\) block of frictionless ice is attached to the other end and rests on the floor. The spring is initially neither stretched nor compressed. A constant horizontal force of 54.0 \(\mathrm{N}\) is then applied to the block, in the direction away from the post. (a) What is the speed of the block when the spring is stretched \(0.400 \mathrm{~m} ?\) (b) At that instant, what are the magnitude and direction of the acceleration of the block?

Your physics book is resting in front of you on a horizontal table in the campus library. You push the book over to your friend, who is seated at the other side of the table, 0.400 m north and 0.300 m east of you. If you push the book in a straight line to your friend, friction does \(-4.8 \mathrm{~J}\) of work on the book. If instead you push the book \(0.400 \mathrm{~m}\) due north and then \(0.300 \mathrm{~m}\) due east, how much work is done by friction?
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