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Chapter 10: Problem 1

A hand exerciser utilizes a coiled spring. A force of \(89.0 \mathrm{N}\) is required to compress the spring by \(0.0191 \mathrm{m} .\) Determine the force needed to compress the spring by \(0.0508 \mathrm{m}\).

Short Answer

Expert verified
The force needed is approximately 236.4 N.

Step by step solution

01

Understand Hooke's Law

Hooke's Law is given by the formula \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement or compression of the spring. Our goal is to find \( k \) using the given force and displacement.
02

Calculate the Spring Constant

We are given that a force of \(89.0 \, \mathrm{N}\) compresses the spring by \(0.0191 \, \mathrm{m}\). According to Hooke's Law, \( k = \frac{F}{x} \). Substitute the given values: \( k = \frac{89.0}{0.0191} \approx 4654.5 \, \mathrm{N/m} \).
03

Calculate the Force for New Compression

Now that we have the spring constant \( k \), we can calculate the force required to compress the spring by a different displacement, \(0.0508 \, \mathrm{m}\). Use Hooke's Law again: \( F = kx \). Substitute \( k = 4654.5 \, \mathrm{N/m} \) and \( x = 0.0508 \, \mathrm{m} \). This gives \( F = 4654.5 \times 0.0508 \approx 236.4 \, \mathrm{N} \).

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Key Concepts

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

Spring Constant
The spring constant, represented by the symbol \( k \), is a fundamental characteristic of a spring. It measures the stiffness of the spring and is defined as the ratio of the force affecting the spring to the displacement caused by it. In simple terms, it tells us how much force is needed to stretch or compress the spring by a unit length. The spring constant is measured in Newtons per meter (N/m).
Essential Points about the Spring Constant:
  • A higher spring constant means a stiffer spring, requiring more force to compress or extend it.
  • A lower spring constant indicates a more flexible spring.
  • In the exercise example, using the formula \( k = \frac{F}{x} \), we calculated the spring constant as \( 4654.5 \, \mathrm{N/m} \). This gives an idea of the stiffness of the hand exerciser's spring.
Force Calculation
Calculating force using Hooke's Law is straightforward once you know the spring constant and the displacement. Hooke's Law is represented by the equation \( F = kx \), where \( F \) is force, \( k \) is the spring constant, and \( x \) is the displacement of the spring. This relationship helps determine how much force is required to achieve a certain level of compression or extension in a spring.
Steps to Calculate Force:
  • First, determine the spring constant \( k \). In this exercise, it's already given as \( 4654.5 \, \mathrm{N/m} \).
  • Next, decide on the displacement \( x \) (how far you wish to compress or stretch the spring).
  • Finally, substitute \( k \) and \( x \) into the equation \( F = kx \) to find the force.
In our example, for a displacement \( x = 0.0508 \, \mathrm{m} \), the calculated force is \( 236.4 \, \mathrm{N} \), demonstrating the amount of effort needed to work the hand exerciser.
Displacement
Displacement in the context of springs refers to the distance a spring is compressed or extended from its equilibrium (resting) position. It is denoted by \( x \) in Hooke's Law. Displacement is an important aspect because how much you compress or stretch the spring directly affects the force required.
Key Considerations about Displacement:
  • Displacement is measured in meters (m) and can be positive or negative, depending on whether the spring is being stretched or compressed.
  • The greater the displacement from the equilibrium position, the greater the force required, as dictated by Hooke's Law formula \( F = kx \).
  • In our exercise, the initial displacement is \( 0.0191 \, \mathrm{m} \), and the new displacement to be considered is \( 0.0508 \, \mathrm{m} \).
Understanding displacement helps you predict the physical effort needed in practical situations involving springs, like exercise equipment or vehicle suspension systems.

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Most popular questions from this chapter

A \(0.60-\mathrm{kg}\) metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to \(0.23 \mathrm{m}\) (relative to its unstrained length), the speed of the sphere decreases from 5.70 to \(4.80 \mathrm{m} / \mathrm{s}\). What is the spring constant of the spring?

To measure the static friction coefficient between a \(1.6-\mathrm{kg}\) block and a vertical wall, the setup shown in the drawing is used. A spring (spring constant \(=510 \mathrm{N} / \mathrm{m}\) ) is attached to the block. Someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. The spring is compressed by \(0.039 \mathrm{m} .\) What is the coefficient of static friction?

A vertical spring (spring constant \(=112 \mathrm{N} / \mathrm{m}\) ) is mounted on the floor. A 0.400-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to \(3.0 \times 10^{-2} \mathrm{N}\). The length and radius of the collagen are, respectively, 2.5 and \(0.091 \mathrm{cm},\) and Young's modulus is \(3.1 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}\). (a) If the stretching obeys Hooke's law, what is the spring constant \(k\) for collagen? (b) How much work is done by the variable force that stretches the collagen? (See Section 6.9 for a discussion of the work done by a variable force.)

A \(1.00 \times 10^{-2}\) -kg bullet is fired horizontally into a 2.50 -kg wooden block attached to one end of a massless horizontal spring \((k=845 \mathrm{N} / \mathrm{m})\). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of \(0.200 \mathrm{m} .\) What is the speed of the bullet?
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