17Calculus Integrals - Work - Hooke's Law
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.
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Here is a quick video clip overview of this.
PatrickJMT - Work [1min-18secs]
video by PatrickJMT |
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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.
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 |
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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 |
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281.25 N-cm
Problem Statement |
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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 |
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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, \(F(x) = x/2 \).
\(\displaystyle{ W = \int_a^b{F(x)~dx} }\) |
\(\displaystyle{ W = \int_{30}^{45}{ x/2 ~dx} }\) |
\(\displaystyle{ W = \left[ \frac{x^2}{4} \right]_{30}^{45} }\) |
\(\displaystyle{ W = \frac{1}{4}[ 45^2 - 30^2 ] = 281.25 }\) N-cm |
Final Answer |
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281.25 N-cm
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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 |
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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 |
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220.5 lbs-in
Problem Statement |
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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 |
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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, \(F(x) = 7x/12 \).
\(\displaystyle{ W = \int_a^b{F(x)~dx} }\) |
\(\displaystyle{ W = \int_{12}^{30}{ 7x/12 ~dx} }\) |
\(\displaystyle{ W = \left[ \frac{7}{24}x^2 \right]_{12}^{30} }\) |
\(\displaystyle{ W = \frac{7}{24}[30^2 - 12^2] = 220.5 }\) lbs-in |
Final Answer |
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220.5 lbs-in
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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 |
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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 |
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1200 video
video by PatrickJMT |
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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 |
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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 |
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1201 video
video by PatrickJMT |
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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 |
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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 |
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1204 video
video by Krista King Math |
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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 |
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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 |
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2.7 foot-pounds
Problem Statement |
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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 |
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1920 video
video by MIP4U |
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Final Answer |
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2.7 foot-pounds
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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 |
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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 |
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37.5 Newtons
Problem Statement |
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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 |
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1919 video
video by MIP4U |
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Final Answer |
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37.5 Newtons
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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 |
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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 |
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2511 video
video by Dr Phil Clark |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
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external links you may find helpful |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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