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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 force-distance.
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 |
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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 \(W=\int_{a}^{b}{F~dx} =\) \(F\int_{a}^{b}{1~dx} =\) \(Fx|_{a}^{b} =\) \(F(b-a)\). Since the distance from a to b is \(d=b-a\), the work is \(W=Fd\).
Result: For a constant force F, an object moved a distance d, the work is \(W=Fd\) |
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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. \(d=\abs{b-a}\). 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 \(W=Fd =\) \((7N)(3m-1m) =\) \(14N\text{-}m\)
Final Answer - \(W=14N\text{-}m\)
Notes -
1. N stands for Newtons and m stands for meters. So the units on the answer are N-m or Newton-meters.
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 |
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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
\(W=\int_{0}^{1}{4x^3~dx} = \) \( \left[ x^4 \right]_{0}^{1} = \) \(1-0 = 1\)
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 force-distance.
Okay, let's work some practice problems before we go on.
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new. |
Work - Practice Problems Conversion |
[1-1205] - [2-1200] - [3-1201] - [4-1204] - [5-1920] - [6-1919] - [7-352] - [8-1202] - [9-1203] |
[10-1206] - [11-1913] - [12-1209] - [13-1914] - [14-1207] - [15-1210] |
Please update your notes to this new numbering system. The display of this conversion information is temporary. |
Calculate the work done by the variable force \(F(x)=\sin(\pi x)\) from \(x=-1\) to \(x=1\).
Problem Statement |
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Calculate the work done by the variable force \(F(x)=\sin(\pi x)\) from \(x=-1\) to \(x=1\).
Solution |
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video by Krista King Math
close 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 |
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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.
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.
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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video by PatrickJMT
close solution |
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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video by PatrickJMT
close solution |
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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video by Krista King Math
close solution |
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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video by MIP4U
Final Answer |
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2.7 foot-pounds |
close solution |
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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video by MIP4U
Final Answer |
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37.5 Newtons |
close solution |
Weight Changing Problems Including Cables, Ropes and Leaking Bags |
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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 \(\int_{0}^{10}{3y~dy} =\) \(\left[3y^2/2\right]_{0}^{10} =\) \(3(100)/2=150\) lbs-ft
Final Answer - \(W=150\) lbs-ft
Find the total work if a rope (4 lbs/ft) hoists a 300lb weight up the side of 500ft building.
Problem Statement |
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Find the total work if a rope (4 lbs/ft) hoists a 300lb weight up the side of 500ft building.
Final Answer |
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total work = 650,000 ft-lbs |
Problem Statement |
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Find the total work if a rope (4 lbs/ft) hoists a 300lb weight up the side of 500ft building.
Solution |
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video by Krista King Math
Final Answer |
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total work = 650,000 ft-lbs |
close solution |
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?
Problem Statement |
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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 |
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video by PatrickJMT
close solution |
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?
Problem Statement |
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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 |
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video by PatrickJMT
close solution |
What is the total work done to lift a 20ft chain that weighs 60lbs, so the top end is 30ft above the ground?
Problem Statement |
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What is the total work done to lift a 20ft chain that weighs 60lbs, so the top end is 30ft above the ground?
Solution |
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video by CalculusSuccess
close solution |
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.
Problem Statement |
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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.
Final Answer |
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3454.5 Joules |
Problem Statement |
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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.
Solution |
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video by MIP4U
Final Answer |
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3454.5 Joules |
close solution |
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.
Problem Statement |
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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 |
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He comes up with a proportion to use as the force that may be difficult to see. We would probably work it differently by saying that the bag starts out at \(144lbs\) but ends the trip at \(72lbs\) (half). We would draw a graph of a straight line (since the leak is a constant rate) between the points \((0,144)\) and \((18,72)\). The equation of that line is the force at any point x, giving us \(F(x)=144-(72)(x/18)\).
video by Dragonometry
close solution |
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 0.2kg/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.
Problem Statement |
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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 0.2kg/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.
Final Answer |
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5233.2 Joules |
Problem Statement |
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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 0.2kg/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.
Solution |
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video by MIP4U
Final Answer |
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5233.2 Joules |
close solution |
Moving Fluid Out of a Tank |
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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.
video by PatrickJMT
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})
Problem Statement |
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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 |
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video by PatrickJMT
close solution |
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.
Problem Statement |
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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 |
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video by Dragonometry
close solution |