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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
236.69 N is needed to compress the spring by 0.0508 m.

Step by step solution

01

Understand Hooke's Law

Hooke's Law describes how springs work and states that the force needed to compress or extend a spring by some distance is proportional to that distance. It can be written as: \[ F = kx \] where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position.
02

Find the Spring Constant

We need to find the spring constant \( k \) using the given information. From the problem, a force \( F = 89.0 \, \mathrm{N} \) compresses the spring by \( x = 0.0191 \, \mathrm{m} \). Using Hooke's Law: \[ k = \frac{F}{x} = \frac{89.0 \, \mathrm{N}}{0.0191 \, \mathrm{m}} \]Calculate this to find \( k \).
03

Calculate the Spring Constant

Calculate the value of \( k \):\[ k = \frac{89.0}{0.0191} \approx 4660.2 \, \mathrm{N/m} \]This is the spring constant of the exerciser.
04

Calculate the Force for New Compression Distance

Now that we have \( k = 4660.2 \, \mathrm{N/m} \), use it to calculate the force needed for a new compression distance of \( x = 0.0508 \, \mathrm{m} \):\[ F = kx = 4660.2 \, \mathrm{N/m} \times 0.0508 \, \mathrm{m} \]Compute this to get the force.
05

Determine the New Force

Perform the multiplication to find the new force:\[ F = 4660.2 \, \mathrm{N/m} \times 0.0508 \, \mathrm{m} \approx 236.69 \, \mathrm{N} \]So, approximately 236.69 N is needed to compress the spring by 0.0508 m.

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Key Concepts

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

Spring Constant
Hooke's Law gives us an essential parameter called the spring constant, denoted as \( k \). This constant tells us how stiff or rigid a spring is.
It measures how much force is needed to compress or extend the spring by a certain distance.
Imagine the spring constant as telling how much a spring resists being stretched or compressed. A higher spring constant means a stiffer spring.
To find the spring constant using Hooke's Law, you use the formula:
  • \( k = \frac{F}{x} \)
  • Here \( F \) is the force applied to the spring, and \( x \) is the distance the spring is compressed or stretched.
For example, if a spring requires 89.0 N of force to be compressed by 0.0191 m, its spring constant \( k \) can be calculated:
\[ k = \frac{89.0 \, \mathrm{N}}{0.0191 \, \mathrm{m}} \approx 4660.2 \, \mathrm{N/m} \].
This result tells us how much force is needed per meter of compression or extension.
Force Calculation
To calculate the force required to compress or stretch a spring, Hooke's Law comes into play.
Once we know the spring constant \( k \), calculating the force for any compression distance \( x \) becomes simple.The formula used is:
  • \( F = kx \)
  • \( F \) represents the force applied, \( k \) is the spring constant, and \( x \) is the compression or stretching distance.
For a displacement of 0.0508 meters, with a spring constant of 4660.2 N/m, we plug these values into the formula:
\[ F = 4660.2 \, \mathrm{N/m} \times 0.0508 \, \mathrm{m} \approx 236.69 \, \mathrm{N} \]
Thus, approximately 236.69 N is needed to compress the spring by 0.0508 m. It's a straightforward multiplication once you know \( k \). Make sure to always double-check your units to ensure they align correctly.
Spring Displacement
Spring displacement refers to how much a spring is compressed or stretched from its original position.
This measurement is crucial as it directly impacts the force needed, as described by Hooke's Law. In real-world applications, knowing the displacement helps in designing systems that rely on spring mechanics, like car suspensions or spring-loaded toys.
To measure displacement:
  • Identify the original, uncompressed length of the spring.
  • Measure how much the spring's length changes when force is applied.
For a spring with a known displacement, you can easily calculate how much force is needed to achieve that change. In our previous example, we used a displacement of 0.0508 m to calculate the required force.
Displacement is often controlled in laboratory settings to test the limits of a spring's elasticity or the durability of a material over repeated use. Understanding this concept is key to mastering topics in physics that involve elasticity and materials science.

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Most popular questions from this chapter

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 174 \(\mathrm{N} / \mathrm{m}\) . The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2 . The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2 .

A vertical spring (spring constant \(=112 \mathrm{N} / \mathrm{m} )\) is mounted on the floor. A \(0.400-\mathrm{kg}\) 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.)

An 11.2 -kg block and a 21.7 -kg block are resting on a horizontal frictionless surface. Between the two is squeezed a spring (spring constant \(=1330 \mathrm{N} / \mathrm{m}\) ). The spring is compresed by 0.141 \(\mathrm{m}\) from its unstrained length and is not attached to either block. With what speed does each block move away after the mechanism keeping the spring squeezed is released and the spring falls away?

A 15.0 -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 \(\mathrm{m} / \mathrm{s}\) in 0.500 \(\mathrm{s}\) . In the process, the spring is stretched by 0.200 \(\mathrm{m}\) . The block is then pulled at a constant speed of 5.00 \(\mathrm{m} / \mathrm{s}\) , during which time the spring is stretched by only 0.0500 \(\mathrm{m} .\) Find \((\mathrm{a})\) the spring constant of the spring and \((\mathrm{b})\) the coefficient of kinetic friction between the block and the table.

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