The equation for work is not that hard. The difficulty arises in determining the integrand, the equation for force. In general, work, \(W\), is the integral of force, \(F(x)\), over a distance from \(x=a\) to \(x=b\). The equation is
\(\displaystyle{W = \int_{a}^{b}{F(x)~dx}}\) with units of forcedistance.
Before we start with specifics of how to calculate work, let's take a few minutes to cover techniques for working word problems.
Word problems are what math students dread the most. I completely understand. I had a lot of difficulty too. However, I found a technique so that I was not only able to successfully work word problems but, eventually, I came to like them and am now able to teach them.
I will tell you up front that figuring out how to work word problems is not easy and takes some independent work on your part to master them. But once you do, you will find them enjoyable and, since so many students struggle with them, most teachers give pretty easy problems, even on exams. So you should be able to breeze your way through them.
First, what doesn't work. Most books try to lump all word problems together and give you general guidelines on how to work them. I have NEVER found that helpful. It wasn't until I was able separate out the different types of word problems, that I came to understand how to work them. Since there are different types of word problems, there are different ways to work them.
Here is what you need to do.
1. Find plenty of problems with worked out solutions. Here are some suggested resources.
 Get a good book with examples and worked out solutions of the type of word problems you are studying. We have posted several suggestions on the books page.
 Check out the solution manual for problems in your textbook.
 17calculus practice problems
2. Once you have a good selection of worked out solutions, go through them carefully and pick up patterns on how they set up the problems, solve them and give the final answer. Pick the ones that are similar to ones in your textbook that you are working on for your class.
3. Key   Group the problems into categories that make sense to you. Some examples might be problems with triangles, problems with right circular cylinders, problems asking you to find areas or volumes. A single problem can go into multiple categories based on configuration or type of question or any other category that makes sense to you.
4. Work the problems yourself before looking at the solutions. Then compare your solutions with the book. Determine what you did wrong and what you need to learn in order to work the problems correctly. At first, this will be slow and painful but once your brain catches on, it will start to be fun. Be patient with yourself, work hard and don't give up. [ In the case of videos, stop the video after the presenter has given the problem statement and work it yourself before watching them solve it. ]
5. Important   Once you have finished a problem, write down the meaning of your answer in words and then reread the problem statement to make sure that your answer is what the problem asked for, including units. For example, I worked a problem about a skydiver and the problem asked for the time it takes for the skydiver to hit the ground after he opens his parachute. When I finished the problem, I had calculated the time that it takes for the skydiver to hit the ground since he jumped out of the plane. When I checked my answer in the back of the book, I was confused until I realized that the number I had was not what the problem asked for. If I had written what my answer means in words and then looked back at the problem, I would have realized right away what I needed to do to finish the problem. Doing this will save you from losing points on homework and exams and it takes only a few seconds.
6. Make sure you understand every single step and, when looking at the solution, figure out why they do things the way they do. If you made a mistake, try to understand what your mistake was and what you need to understand in order to not make the same mistake again. [ Also remember that no textbook or video is always 100% correct. If you can not figure out your mistake, find someone to ask and see if the solution manual is incorrect. ]
7. Pick up patterns and general ideas from each group of problems by working the same type of problems all together. Don't jump around to different types. Stay with one type for several problems. I won't tell you exactly how many. You need to determine that by how difficult the problems are, how well you think you understand the current type, how much time you have and how well you want to do on your homework and exams. Sometimes you can go on after working 5 of the same type, sometimes it takes 10 or more.
8. Find a friend to work with and go over the problems with them AFTER you have worked them on your own. Remember, at exam time you will be on your own. So don't rely on someone else too much. If you know more than the other person, explain your work to them. Communicating your work to someone else helps you understand it better. If you know less, ask lots of questions and ask them to explain their solution to you.
9. Do NOT do shortcuts. Shortcuts are good AFTER you have learned the material, not while you are learning the material. Do it the long way for a while until you are know it really well.
10. 2nd Key   Do not just look at the solutions or watch someone else work the problems. You need to get out a pencil and paper and work them yourself. You are going to get frustrated. You are going to want to quit, but don't quit. Use that feeling to motivate yourself and show yourself that you can do it. It feels great to master something that is difficult. If you have never pushed through something difficult before, try it now. It is not easy but it is worth it. I know because I went through this same process myself.
11. Finally, do not skip ANYTHING and NEVER GIVE UP. Make sure you understand every single step in every single problem. Here's why: Chances are, if you skip something, it will show up on an exam precisely because the part you don't understand is probably the most difficult part of the problem and teachers expect you to skip it. So they put it on exams to see if you understand the difficult parts.
So far, I have found that implementing these ideas as the best way to figure out how to work word problems. There are tons of general guidelines in books (most likely in your textbook too) that never really helped me. Give this technique a try. Remember, you are now in charge of your own learning. No one is going to help you from here on out. You need to do it.
As described in the panel above, it helps to arrange problems in groups and then work several problems in a row in the same group. We have grouped the problems according to type of problem usually found in physics. However, if you need to, arrange them differently so that they make sense to you.
Constant Force
Let's start with something you are probably familiar with, a constant force. If you have seen work before in physics, you were probably told that work is force times distance. This is true when the force is constant. Let's see why.
As we said above, the equation for work is \(W=\int_{a}^{b}{F(x)~dx}\) when an object is moved from the point \(x=a\) to \(x=b\). When the force is constant, then we have \(W=\int_{a}^{b}{F~dx} =\) \(F\int_{a}^{b}{1~dx} =\) \(Fx_{a}^{b} =\) \(F(ba)\). Since the distance from a to b is \(d=ba\), the work is \(W=Fd\).
Result: For a constant force F, an object moved a distance d, the work is \(W=Fd\) 

Note: We are assuming in the above discussion that \(b > a\). To be completely accurate and to cover the case when b may not be greater than a, we need absolute value signs on d, i.e. \(d=\abs{ba}\). Okay, let's do an example.
Example  If a force of 7 N moves an object from x=1 m to x=3 m, what is the work done by the force?
Solution  Since the force is constant, we know that the work is \(W=Fd =\) \((7N)(3m1m) =\) \(14N\text{}m\)
Final Answer  \(W=14N\text{}m\)
Notes 
1. N stands for Newtons and m stands for meters. So the units on the answer are Nm or Newtonmeters.
2. In the example, we noticed that the force is a constant. So we jumped directly to the shortcut formula \(W=Fd\). Of course, you could have done this by using the integral and ended with the same answer. Check with your instructor to see what they expect.
The rest of this page and most of the problems you will encounter involve forces that are not constant. First, let's look at the case when the force is actually given to you.
Given a General Force
If you are given a force, all you need to do is write down the integral and evaluate it. Before we get into the details, let's watch a quick explanation video of work.
video by Michel vanBiezen 

Example  When a force \(F(x)=4x^3\) moves an object from x=0 to x=1, calculate the work done by the force.
Solution  Our equation is \(W=\int_{a}^{b}{F(x)~dx}\). Using the information in the problem statement gives us
\(W=\int_{0}^{1}{4x^3~dx} = \) \( \left[ x^4 \right]_{0}^{1} = \) \(10 = 1\)
Final Answer  \(W=1\)
Note  Notice that since we are not given units in the problem statement, we do not have units on our answer. This is okay as long as we remember that the work has units forcedistance.
Okay, let's work some practice problems before we go on.
Practice
Calculate the work done by \(F(x)=\sin(\pi x)\) from \(x=1\) to \(x=1\).
Problem Statement 

Calculate the work done by \(F(x)=\sin(\pi x)\) from \(x=1\) to \(x=1\).
Solution 

video by Krista King Math 

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A force of \( F(x) = xe^{2x^2} + 6x  2 \) acts on an object. What is the work required to move the object from \( x = 1 \) to \( x = 4 \)?
Problem Statement 

A force of \( F(x) = xe^{2x^2} + 6x  2 \) acts on an object. What is the work required to move the object from \( x = 1 \) to \( x = 4 \)?
Final Answer 

\(\displaystyle{ 39 + \frac{e^{30}1}{4e^{32}} \approx 39.03}\)
Problem Statement 

A force of \( F(x) = xe^{2x^2} + 6x  2 \) acts on an object. What is the work required to move the object from \( x = 1 \) to \( x = 4 \)?
Solution 

Work is calculated using the equation \(\displaystyle{ W = \int_a^b{F(x)~dx} }\).
There is not much to this problem since everything is given to us. We just need to set up the integral and evaluate it.
\(\displaystyle{ W = \int_a^b{F(x)~dx} }\) 
\(\displaystyle{ W = \int_1^4{ xe^{2x^2} + 6x  2 ~dx} }\) 
Let's work on the first term separately. 
\(\displaystyle{ \int{ xe^{2x^2} ~dx} }\) 
\(u=2x^2 \) 
\( du=4x~dx \to du/4 = x~dx\) 
\(\displaystyle{ (1/4) \int{ e^u ~du} }\) 
\(\displaystyle{ (1/4) e^u \to (1/4) e^{2x^2} }\) 
\(\displaystyle{ W = \left[ \frac{1}{4}e^{2x^2} +3x^2  2x \right]_1^4 }\) 
\(\displaystyle{ \left[ \frac{1}{4}e^{32} + 48  8 \right]  \left[ \frac{1}{4}e^{2} + 3  2 \right] }\) 
\(\displaystyle{ 39 + \frac{e^{30}1}{4e^{32}} \approx 39.03}\) 
There were no units given in the problem statement, so none are required in the answer.
Final Answer 

\(\displaystyle{ 39 + \frac{e^{30}1}{4e^{32}} \approx 39.03}\) 
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A force of \( F(x) = 4\cos(2x)  7\sin(x/2) \) where x is in meters, acts on an object. What is the work required to move the object 10 meters to the right of x=2?
Problem Statement 

A force of \( F(x) = 4\cos(2x)  7\sin(x/2) \) where x is in meters, acts on an object. What is the work required to move the object 10 meters to the right of x=2?
Hint 

Although not stated in the problem, we will assume that the initial position is x=2.
Problem Statement 

A force of \( F(x) = 4\cos(2x)  7\sin(x/2) \) where x is in meters, acts on an object. What is the work required to move the object 10 meters to the right of x=2?
Final Answer 

\( 2[ \sin(24)\sin(4) ] + \) \(14[ \cos(6)  \cos(1) ] \approx 5.58 \) Nm
Problem Statement 

A force of \( F(x) = 4\cos(2x)  7\sin(x/2) \) where x is in meters, acts on an object. What is the work required to move the object 10 meters to the right of x=2?
Hint 

Although not stated in the problem, we will assume that the initial position is x=2.
Solution 

Work is calculated using the equation \(\displaystyle{ W = \int_a^b{F(x)~dx} }\).
\(\displaystyle{ W = \int_a^b{F(x)~dx} }\) 
\(\displaystyle{ W = \int_2^{12}{ 4\cos(2x)  7\sin(x/2) ~dx} }\) 
\(\displaystyle{ W = \left[ 2\sin(2x) + 14\cos(x/2) \right]_2^{12} }\) 
\( 2\sin(24) + 14\cos(6)  2\sin(4)  \) \( 14\cos(1) \) 
\( 2[ \sin(24)\sin(4) ] + \) \( 14[ \cos(6)  \cos(1) ] \approx 5.58 \) 
Since the distance is in meters, we assume the force is in Newtons, giving the units of work as Newtonmeters.
Final Answer 

\( 2[ \sin(24)\sin(4) ] + \) \(14[ \cos(6)  \cos(1) ] \approx 5.58 \) Nm 
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A force of \( F(x) = \sin(x)e^{\cos x}  4x + 1 \), x is in meters, acts on an object. What is the work required to move the object 6.5 meters to the left of x=9?
Problem Statement 

A force of \( F(x) = \sin(x)e^{\cos x}  4x + 1 \), x is in meters, acts on an object. What is the work required to move the object 6.5 meters to the left of x=9?
Hint 

We are not given an initial position but we will assume that we start at x=9. Subtracting 6.5 from 9 gives us the final position of x=2.5.
Problem Statement 

A force of \( F(x) = \sin(x)e^{\cos x}  4x + 1 \), x is in meters, acts on an object. What is the work required to move the object 6.5 meters to the left of x=9?
Final Answer 

\( W = 143  e^{\cos 2.5} + e^{\cos 9} \approx 142.953 \) Nm
Problem Statement 

A force of \( F(x) = \sin(x)e^{\cos x}  4x + 1 \), x is in meters, acts on an object. What is the work required to move the object 6.5 meters to the left of x=9?
Hint 

We are not given an initial position but we will assume that we start at x=9. Subtracting 6.5 from 9 gives us the final position of x=2.5.
Solution 

Work is calculated using the equation \(\displaystyle{ W = \int_a^b{F(x)~dx} }\).
\(\displaystyle{ W = \int_a^b{F(x)~dx} }\) 
\(\displaystyle{ W = \int_{9}^{2.5}{ \sin(x)e^{\cos x}  4x + 1 ~dx} }\) 
\(\displaystyle{ W = [ e^{\cos x}  2x^2 + x ]_9^{2.5} }\) 
\( W = [ e^{\cos 2.5}  2(2.5)^2 + 2.5 ]  \) \( [ e^{\cos 9}  2(9)^2 + 9 ] \) 
\( W = 143  e^{\cos 2.5} + e^{\cos 9} \approx 142.953 \) 
Since the distance is in meters, we assume the force is in Newtons, giving the units of work as Newtonmeters.
Final Answer 

\( W = 143  e^{\cos 2.5} + e^{\cos 9} \approx 142.953 \) Nm 
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Okay, so that was not too bad. But from now on you will probably need to come up with the equation for force. This is where we need to divide the problems into different types so that you can start to see patterns for how to solve them. Let's start with a very common type of problem involving springs called Hooke's Law.
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related topics 

external links you may find helpful 
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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