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Applied Integration  Work 

on this page: ► constant force ► general force ► Hooke's Law ► weight changing problems ► moving fluid 
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. 
Guidelines For Working Word ProblemsWord 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.


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. Here is an example.
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 1 

Calculate the work done by the variable force \(F(x)=\sin(\pi x)\) from \(x=1\) to \(x=1\). 
solution 
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.
Hooke's Law  Spring Problems 

Okay, so 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 a 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 [1min18secs]  
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 2 

A force of 10lb is required to hold a spring stretched 4in beyond its natural length. How much work (in footpounds) is done in stretching it from its natural length to 6in beyond its natural length? 
solution 
Practice 3 

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 
Practice 4 

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 
Practice 5  

A force of 3lbs is required to hold a spring stretched 0.2 feet beyond its natural length. How much work (in footpounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?  
answer 
solution 
Practice 6  

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)?  
answer 
solution 
Weight Changing Problems Including Cables, Ropes and Leaking Bags 

These type of problems are interesting in that we need to take into account the weight of the cable or rope itself or the changing weight of the bag of sand or water or whatever is in the bag. In problems you've probably seen before, lifting a weight alone is just force times distance. But if you take into account the changing weight, with each small distance you lift, there is less weight. So the force required is not constant, meaning we need to use an integral to calculate work. So here is how we work these problems.
First of all let me say that drawing diagrams for these problems so that you understand what is going is critical. Otherwise it will be nearly impossible to solve these kinds of problems because of the many variations that are possible. We will cover the case where the force required to lift the object is changing. Some problems add static weights to the problem or raise the object beyond a point where the force is changing. These kind of problems are included in the practice problems.
Once you have drawn all the diagrams you need for the problem and you think you understand what is going on, you need to write an equation for the changing force. Some instructors set up an infinite sum then convert to an integral. Several of the practice problems do that too. However, I am going to show you another way to think about it that I think will help you. Let's look at a chain example.
Example  A 10ft chain that weighs 30lbs is initially laying on the ground. Calculate the work to lift one end of the chain 10ft into the air.
Solution  Let's think about what is going on. First, we pick up one end of the chain and start lifting. As we get further up in the air, the weight we are lifting is getting heavier and heavier until we have all 10ft hanging and we are holding 30lbs of chain.
Some instructors have you think about each link in the chain and how much work it takes to raise each link. I prefer to think about only the link that I am holding and, depending on the height of that one link, it's weight is getting heavier. So at 1ft, I am holding 3lbs (30lbs/10ft * 1ft = 3lbs). As I raise it higher and higher, that one link is getting heavier and heavier, so it takes more work to lift the last foot of the chain than it does to raise it the first foot. Do you have the idea in your head?
So, at each position y, I need to exert a force of \(30lbs/10ft \cdot y = 3y\). The quantity 30lbs/10ft = 3 lbs/ft is the density of the chain. The more chain there is to raise, the more force required.
So, if we go from \(y=0\)ft to \(y=10\)ft, the work required is \(\int_{0}^{10}{3y~dy} =\) \(\left[3y^2/2\right]_{0}^{10} =\) \(3(100)/2=150\) lbsft
Final Answer  \(W=150\) lbsft
Practice 7  

Find the total work if a rope (4 lbs/ft) hoists a 300lb weight up the side of 500ft building.  
answer 
solution 
Practice 8 

A heavy rope, 60ft long, weighs 2lbs/ft and hangs over the edge of a building 120ft high. How much work is required to pull the entire rope to the top? 
solution 
Practice 9 

A heavy rope, 60ft long, weighs 2lbs/ft and hangs over the edge of a building 120ft high. How much work is required to pull half of the rope to the top? 
solution 
Practice 10 

What is the total work done to lift a 20ft chain that weighs 60lbs, so the top end is 30ft above the ground? 
solution 
Practice 11  

A leaky bucket full of sand has a mass of 25 kg. The bucket is to be lifted to the top of a building 15 meters tall by a rope of negligible weight. Every meter the bucket is lifted results in a loss of 0.2 kg of sand. Find the work done lifting the bucket to the top of the building.  
answer 
solution 
Practice 12 

We are lifting a leaky bag sand that initially weighs 144lbs. The bag is half empty at 18ft. Calculate the work required to lift the bag of sand initially on the ground to 18ft above the ground. Assume the leak is a constant rate. 
solution 
Practice 13  

A bucket with a mass of 30 kg when filled with sand needs to be lifted to the top of a 20 meter tall building. We have a rope that has a mass of 0.2 kg/m that takes 1 meter to secure to the bucket. Once the bucket reaches the top of the building it has a mass of only 19 kg because there is a hole in the bottom and sand was leaking out at a constant rate while it was being lifted to the top of the building. Find the work done lifting the bucket, sand and rope to the top of the building.  
answer 
solution 
Moving Fluid Out of a Tank 

The idea of these problems is to move fluid out of a tank by raising it against gravity. The hard part is how to set up the differential \(dW\).
Here is a quick video clip giving an overview.
PatrickJMT  Moving Fluid  
Practice 14 

We have a 10ft long trough with cross section as a triangle with base 3ft and height 3ft (point of triangle at the bottom of the tank). Calculate the work required to drain the entire tank by pumping the water out of a pipe 2ft above the top of the tank. (density of water is 62.5lb/ft^{3}) 
solution 
Practice 15 

We have a cylindrical tank, half full of water, which is 10m high and radius 5m. Calculate the work required to pump the water out of the tank to a level 4m above the top of the tank. Use 9800 N/m^{3} for the weight of water. 
solution 