\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Integrals - Work - Hooke's Law

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

We have the general equation for work as \( \int_{a}^{b}{F(x)~dx} \). When we have a spring problem, Hooke's Law says that the force equation is \(F(x)=kx\), where k is a constant called the spring constant. The spring constant is an attribute of the spring itself and it is a measure of the 'stiffness' of the spring.

Here is a quick video clip overview of this.

PatrickJMT - Work [1min-18secs]

video by PatrickJMT

Natural Position - In that video clip, he mentioned something about the natural position of the spring. What is that? Well, imagine that you have a vertical wall and an object sitting on the floor near the wall. Now, imagine that there is a spring between them, one end attached to the wall, the other attached to the object and that the spring is exactly parallel to the floor. Now, on the floor there is very, very cold ice, maybe even dry ice, so that the object is on the ice and it will move with even the slightest push. The natural position of the spring is when we have this configuration and the object is not moving, i.e. the spring is neither pushing nor pulling on the object when it is at rest. The object is just sitting there with no interference from anything or anyone else. That is the natural position of the spring.

Value of k - Now, if you are given the value of k, then the problem is easy to set up. Your force is \(F(x)=kx\). Unfortunately, most of the time, you will not be given k outright. You usually need to calculate it. If you do, you will usually find two situations in the problem statement. Look for a combination of static distance and force numbers to use in the equation \(F(x) = kx\). Then use the second situation in the work equation.

For example, you may have a part of the problem that says something like, "it takes a force of 2N to extend the spring 4cm from it's natural position". Since \(F=kx\), \(2N=k(4cm) \to\) \(k=(1/2)N/cm\).

That's about as hard as these types of problems get. But you need to work several of them that are worded differently in order to be able to work them easily on your exam. Try several practice problems before starting your homework.

You Can Have an Amazing Memory: Learn Life-Changing Techniques and Tips from the Memory Maestro

Practice

A spring has a natural length of 25 cm and a force of 3.5 N is required to stretch and hold the spring to a length of 32 cm. What is the work required to stretch the spring from 30 cm to 45 cm?

Problem Statement

A spring has a natural length of 25 cm and a force of 3.5 N is required to stretch and hold the spring to a length of 32 cm. What is the work required to stretch the spring from 30 cm to 45 cm?

Final Answer

93.75 N-cm

Problem Statement

A spring has a natural length of 25 cm and a force of 3.5 N is required to stretch and hold the spring to a length of 32 cm. What is the work required to stretch the spring from 30 cm to 45 cm?

Solution

Work is calculated using the equation \(\displaystyle{ W = \int_a^b{F(x)~dx} }\).
The force is from Hooke's Law, \(F(x) = kx\). First, we need to calculate \(k\).
We know a force of 3.5N will stretch the spring from 25cm to 32cm. So \( 3.5 = k(32-25) \to k=1/2\).
Now we have what we need, \(F(x) = x/2 \).

\(\displaystyle{ W = \int_a^b{F(x)~dx} }\)

\(F(x) = x/2\) and the limits of integration are with respect to the natural length. So the limits of integration are from 5cm to 20cm.

\(\displaystyle{ W = \int_{5}^{20}{ x/2 ~dx} }\)

\(\displaystyle{ W = \left[ \frac{x^2}{4} \right]_{5}^{20} }\)

\(\displaystyle{ W = \frac{1}{4}[ 20^2 - 5^2 ] = 93.75 }\) N-cm

Final Answer

93.75 N-cm

Log in to rate this practice problem and to see it's current rating.

A spring has a natural length of 9 inches and a force of 7 lbs is required to stretch and hold the spring to a length of 21 inches. What is the work required to stretch the spring from 12 inches to 30 inches?

Problem Statement

A spring has a natural length of 9 inches and a force of 7 lbs is required to stretch and hold the spring to a length of 21 inches. What is the work required to stretch the spring from 12 inches to 30 inches?

Final Answer

126 lbs-in

Problem Statement

A spring has a natural length of 9 inches and a force of 7 lbs is required to stretch and hold the spring to a length of 21 inches. What is the work required to stretch the spring from 12 inches to 30 inches?

Solution

Work is calculated using the equation \(\displaystyle{ W = \int_a^b{F(x)~dx} }\).
The force is from Hooke's Law, \(F(x) = kx\). First, we need to calculate \(k\).
We know a force of 7lbs will stretch the spring from 9in to 21in. So \( 7 = k(21-9) \to k=7/12\) lbs/in.
Now we have what we need, \(F(x) = 7x/12 \).

\(\displaystyle{ W = \int_a^b{F(x)~dx} }\)

So we found \(F(x) = 7x/12 \). The limits of integration need to be with respect to the natural length. So the limits are from 3in to 21in.

\(\displaystyle{ W = \int_{3}^{21}{ 7x/12 ~dx} }\)

\(\displaystyle{ W = \left[ \frac{7}{24}x^2 \right]_{3}^{21} }\)

\(\displaystyle{ W = \frac{7}{24}[21^2 - 3^2] = 126 }\) lbs-in

Final Answer

126 lbs-in

Log in to rate this practice problem and to see it's current rating.

A force of 10lb is required to hold a spring stretched 4in beyond its natural length. How much work (in foot-pounds) is done in stretching it from its natural length to 6in beyond its natural length?

Problem Statement

A force of 10lb is required to hold a spring stretched 4in beyond its natural length. How much work (in foot-pounds) is done in stretching it from its natural length to 6in beyond its natural length?

Solution

PatrickJMT - 1200 video solution

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

Suppose 2J of work is needed to stretch a spring from its natural length of 30cm to a length of 42cm. How much work is required to stretch it from 35cm to 40cm?

Problem Statement

Suppose 2J of work is needed to stretch a spring from its natural length of 30cm to a length of 42cm. How much work is required to stretch it from 35cm to 40cm?

Solution

PatrickJMT - 1201 video solution

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

A spring has a natural length of 2ft and a force of 15 pounds is required to hold it compressed at a length of 18in. How much work is done in stretching the spring from its natural length to a length of 3ft?

Problem Statement

A spring has a natural length of 2ft and a force of 15 pounds is required to hold it compressed at a length of 18in. How much work is done in stretching the spring from its natural length to a length of 3ft?

Solution

Krista King Math - 1204 video solution

video by Krista King Math

Log in to rate this practice problem and to see it's current rating.

A force of 3lbs is required to hold a spring stretched 0.2 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?

Problem Statement

A force of 3lbs is required to hold a spring stretched 0.2 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?

Final Answer

2.7 foot-pounds

Problem Statement

A force of 3lbs is required to hold a spring stretched 0.2 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?

Solution

MIP4U - 1920 video solution

video by MIP4U

Final Answer

2.7 foot-pounds

Log in to rate this practice problem and to see it's current rating.

Work of 3J is done in stretching a spring from its natural length of 16cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (16cm)?

Problem Statement

Work of 3J is done in stretching a spring from its natural length of 16cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (16cm)?

Final Answer

37.5 Newtons

Problem Statement

Work of 3J is done in stretching a spring from its natural length of 16cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (16cm)?

Solution

MIP4U - 1919 video solution

video by MIP4U

Final Answer

37.5 Newtons

Log in to rate this practice problem and to see it's current rating.

A spring has a natural length of 20 cm. If a 25 Newton force is required to keep it stretched to 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

Problem Statement

A spring has a natural length of 20 cm. If a 25 Newton force is required to keep it stretched to 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

Solution

Dr Phil Clark - 2511 video solution

video by Dr Phil Clark

Log in to rate this practice problem and to see it's current rating.

Really UNDERSTAND Calculus

Log in to rate this page and to see it's current rating.

Topics You Need To Understand For This Page

Related Topics and Links

To bookmark this page and practice problems, log in to your account or set up a free account.

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

math and science learning techniques

Try Audible Premium Plus and Get Up to Two Free Audiobooks

As an Amazon Associate I earn from qualifying purchases.

I recently started a Patreon account to help defray the expenses associated with this site. To keep this site free, please consider supporting me.

Support 17Calculus on Patreon

Practice Search

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2022 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics