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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 square plate is \(1.0 \times 10^{-2} \mathrm{m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{m}\) on a side, and has a mass of \(7.2 \times 10^{-2}\) kg. The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is 0.91 . A force is applied to the top of the plate, as in Figure \(10.29 a .\) Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10.29 b\) ) that can be created by the applied force just before the plate begins to move.

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by \(3.00 \mathrm{m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as the height of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of \(26.8 \mathrm{m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

An archer, about to shoot an arrow, is applying a force of \(+240 \mathrm{N}\) to a drawn bowstring. The bow behaves like an ideal spring whose spring constant is \(480 \mathrm{N} / \mathrm{m} .\) What is the displacement of the bowstring?

A person who weighs \(670 \mathrm{N}\) steps onto a spring scale in the bathroom, and the spring compresses by \(0.79 \mathrm{cm} .\) (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{cm} ?\)

A spring stretches by \(0.018 \mathrm{m}\) when a \(2.8-\mathrm{kg}\) object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)
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