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Energy -- Work

Module by: R.G. (Dick) Baldwin. E-mail the author

Summary: This module explains energy and work in a format that is accessible to blind students.

Preface

General

This module is part of a collection (see http://cnx.org/content/col11294/latest/ ) of modules designed to make physics concepts accessible to blind students. The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.

This module explains energy and work in a format that is accessible to blind students.

Prerequisites

In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:

The minimum prerequisites for understanding the material in these modules include:

Supplemental material

I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .

General background information

I will explain some of the background material regarding work in this section and show how to work problems involving work later .

An assumption of constant velocity

In order to simplify the solutions to the problems, many of the example problems in this module will make the assumption that an object is moving with constant velocity. Let's review what that means in terms of what you have learned from earlier modules.

Unbalanced forces cause acceleration

We know that whenever unbalanced forces are applied to an object, that object will accelerate. What does it mean, therefore, when we make a statement like "a force of 1 newton to the right was applied to an object to cause that object to move 1 meter to the right at a constant velocity"?

Conditions for constant velocity

We know that the statement means at least the following things:

  • At some point in the past, there was an unbalanced force applied to the object to cause it to begin moving to the right. That force is ignored by the statement of the problem.
  • At the beginning of the time frame covered by the problem, the object was in equilibrium, moving at a constant velocity to the right.
  • During the time frame of the problem, there must be a 1 newton balancing force being exerted to the left to prevent the 1 newton force being applied to the right from causing the object to accelerate towards the right. That force is also ignored by the statement of the problem.

Use your graph board for sketches

As a sighted person, I often find it useful to make sketches of the physical scenarios for problems of the types that I will discuss in this module.

You may also find it useful to use your graph board to create "sketches" of the physical scenarios in order to better understand the solutions to the problems.

Work in everyday life

You hear the word work in many situations in everyday life. For example, someone might say "My gasoline bill is high because I have to drive a long way to work every day."

Someone else might say "She works very hard at being popular."

Or, you might hear someone say "It's going to take a lot of work to lift that heavy sofa into the back of that truck."

The latter expression comes closest to describing what is mean by work in physics.

Work in physics

According to College Physics by Mendenhall et al

If a body on which a force F is acting is displaced by a distance d, then the work done by the force is F*d*cos(theta), where theta is the angle between the path d and the line of action of F.

Thus, the term work has a very precise meaning in physics. In equation form, we can represent work as

W = F * d * cos(theta)

where

  • W represents work
  • F represents a force that causes the displacement of an object
  • d represents the magnitude of the displacement
  • theta is the angle between the line of action of the force and the direction of the displacement

Three key ingredients

According to the Physics Classroom (see http://www.physicsclassroom.com/Class/energy/u5l1a.cfm ), there are three key ingredients to work: force , displacement , and cause .

In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement.

Commonplace examples of work

Examples of work are commonplace in everyday life:

  • A bulldozer pushing dirt. The bulldozer exerts a horizontal force to move the dirt.
  • An elevator lifting people from the first to the tenth floor. The cable on the elevator exerts an upward force to move the people upward by approximately 110 feet.
  • You lifting your backpack full of books onto your back. Your muscles exert a force to raise several heavy books, possibly displacing them both vertically and horizontally at the same time.

However, some things that we might initially think of as work don't really qualify as work according to the physics definition of work.

Suzie and the heavy trunk

Your college dorm is equipped with a rope on a pulley that can be used to lift heavy objects and swing them into a window on the second floor. Suzie has used the pulley and the rope to lift her heavy trunk up to the level of the window. She asks you to hold the rope and maintain the trunk at that level while she goes up the stairs to pull it in the window.

Suzie likes to gossip

Suzie, however, really likes to gossip. On the way up to the window, she meets some friends and starts talking, leaving you holding onto the rope. After about 30 minutes, you become really tired and frustrated and you lower the trunk back to ground level.

Did you do any work on the trunk?

The answer is that you did not do any work on the trunk. Although you became very hot and tired standing there and holding the rope, you did not cause a displacement of the trunk. For 30 minutes, you simply held the rope causing the trunk to remain at the same location. At the end of that period, you carefully allowed the rope to slip through your hands while the force of gravity caused a downward displacement on the trunk.

Was any work done on the trunk?

Was any work done on the trunk from the time Suzie handed you the rope until the trunk was back on the ground?

Yes, the force of gravity caused the height of the trunk to be displaced from the level of the second floor window downward to the ground, so the force of gravity did work on the trunk.

Was any work done on the trunk from the time that Suzie started raising it until it was back on the ground?

The answer is no. The net displacement of the trunk over that time period was zero. If the displacement was zero, no work could have been done. In this case, the positive work done by Suzie in lifting the trunk to the second floor window was canceled out by the negative work done by gravity when you allowed the trunk to return to the ground.

Measurement of work can depend on the time interval

As you can see, therefore, the amount of work done in some cases can depend on the time interval over which the measurement is made.

Also, as you can see, certain activities (such as holding a rope to keep a trunk from falling) can make you very tired even though you are not doing any work.

Mathematical representation of work

As stated earlier, we can represent work as

W = F * d * cos(theta)

where

  • F represents a force being applied to an object
  • d represents the displacement of the object
  • theta is the angle between the displacement path d and the line of action of the force.

Work is a scalar quantity

Although both force and displacement are vector quantities, the product of those two vectors with the cosine of an angle produces a scalar quantity. Work has a magnitude but no direction. The product is often called a scalar product or a dot product.

The scalar or dot product

The scalar product (or dot product ) of two vectors is defined by the equation

vectorA * vectorB * cos(theta)

where theta is the angle between the two vectors when they are drawn tail to tail.

The special name and notation are used because this pattern occurs often in physics and mathematics.

Units of work

The SI unit of work (and energy) is the newton-meter (N*m), which is given the name joule (symbol: J), after the English physicist James Prescott Joule. One joule is equivalent to one newton of force causing a displacement of one meter.

A stubborn dog

Pretend that you have a dog that is very stubborn and just wants to sleep. You pull on the dog's leash and cause the dog to slide 2 meters across the floor. In this case, the displacement is horizontal and you are pulling at an angle of 30 degrees above the horizontal.

The work done on the dog would be the product of the horizontal component of the force that you exert on the leash (F*cos(theta)) and the displacement distance of 2 meters. If you were pulling with a constant horizontal force component of 3 newtons, then you would have done 2*3=6 joules of work.

A stubborn pig

A very large pig has escaped from its pen. You have a rope tied around the pig's neck and you are trying to pull the pig back into the pen. The rope is horizontal and you pull with a steady force of 3 newtons.

While you are pulling, however, the pig is backing up (actually pulling you forward away from the pen), and travels 2 meters in the direction opposite the direction that you are pulling.

In this case, the angle between the line of action of the force that you are exerting and the direction of the pig's displacement is 180 degrees. Since the cosine of 180 degrees is -1, the product of the force, displacement, and cosine of 180 degrees would be -6. Therefore, your efforts to move the pig toward the pen are unsuccessful and you have done -6 joules of work on the pig.

Therefore, work can be either positive or negative.

More on that stubborn pig

The very large pig has decided to lay down on a hydraulic lift on the back of a truck and someone accidentally flipped the switch causing the hydraulic lift to start moving up.

You still have a rope tied around the pig's neck and you attempt to pull the pig off the lift to prevent it from getting hurt. You pull as hard as you can on the rope but the pig doesn't budge. The lift goes up two meters to the height of the truck bed and turns itself off automatically with the pig still laying on the lift.

The angle is 90 degrees

In this case, the angle between the line of action of the force that you exert and the direction of displacement of the pig is 90 degrees. The cosine of 90 degrees is 0. Therefore, the product of the force, the displacement, and the cosine of 90 degrees would be 0.

While you might be ready for a rest after exerting all of that effort trying to pull the pig off the lift, you actually haven't done any work on the pig.

Therefore, work can even be zero if the line of action of the force is perpendicular to the displacement of the object. In this case, work did get done to raise the pig 2 meters off the ground, but you didn't contribute anything to that work. The work was done by the mechanical system that operates the hydraulic lift on the truck.

To do any work, a force must cause a displacement.

More on the stubborn dog

Let's go back and think a little more about the stubborn dog. Assume that you exert a force of 3.46 newtons on the leash and the leash makes an angle of 30 degrees with the horizontal floor. This causes the dog to slide 2 meters across the floor, but doesn't lift the dog off the floor. What is the work that you do on the dog?

The horizontal component Fx of the force F is given by

Fx = F*cos(30 degrees), or

Fx = 3.46*cos(30 degrees)

Enter the following into the Google calculator search box:

3.46*cos(30 degrees)

and with rounding, you will learn that the horizontal component of the force that you exerted on the leash is

3 newtons.

How much work was done?

The work done on the dog is the product of the horizontal component of the force that you exert on the leash and the horizontal displacement distance of 2 meters.

Assuming that you were pulling the leash with a constant force of 3.46 newtons, at an angle of 30 degrees relative to the horizontal floor, you would do 2*3=6 joules of work on the dog.

Be careful how you handle the angle theta

The angle theta is the angle between the line of action of the force and the direction of the displacement. It is only because I set the scenario up that way that the 30-degree angle in the previous stubborn-dog scenario is 30 degrees relative to the horizontal.

In that scenario, the line of action of the 3.46-newton force was at 30 degrees relative to the horizontal, but the displacement was horizontal. Therefore, it was necessary to take the angle into account when computing the work done on the dog. In other words, the work done on the dog was:

W = 3.46N * 2m * cos(30 degrees) = 6 joules

One more discussion about the stubborn dog

Assume that instead of pulling the dog across a horizontal floor, you are pulling the dog up a ramp that has an angle of 30 degrees relative to the horizontal.

In this case, you pull on the leash at an angle of 30 degrees relative to the horizontal, which is parallel to the ramp. Once again, you pull with a force of 3.46 newtons. This time, you cause the dog to slide 2 meters up the ramp.

In this case, the angle between the direction of the displacement and the line of action of the force is 0 degrees. The cosine of 0 degrees is +1. Therefore, the work done on the dog is

(3.46 N) * 2 m * cos(0 degrees) = 6.92 joules

(You may recognize the format of some of the mathematical expressions in the above paragraphs as being the format of the output of the Google calculator.)

Sample calculations

I will present and explain several scenarios in this section involving force and work.

In some cases, I will use the Google calculator to perform the arithmetic, and the format of the mathematical expressions will be the format of the output from the Google calculator (except that I will usually discard some of the digits to the right of the decimal point).

Single force on a body

A horizontal force

A horizontal force of 10 N towards the right is applied to a 15-kg body causing the body to move 5 m to the right at constant speed. How much work is done on the body?

Answer:

(10 newtons) * 5 meters = 50 joules

Note that the mass of the object is superfluous in this problem.

A force at 30 degrees

A force of 10 N is applied to a 15-kg body at an angle of 30 degrees relative to the horizontal causing the body to move 5 m to the right at constant speed. How much work is done on the body?

Answer:

(10 N) * 5 m * cos(30 degrees) = 43.3 joules

Once again, the mass of the object is superfluous.

A vertical force

The mass is not superfluous for this scenario.

A force is applied to a 10-kg body at an angle of 90 degrees relative to the horizontal causing the body to move 5 m straight up at constant velocity. How much work is done on the body?

Answer:

Weight = (10 kg) * 9.81 * (m / (s^2)) = 98.1 newtons

Under the constant velocity assumption explained earlier, an upward force equal to the weight of the object will cause the object to continue moving in an upward direction, and eventually move 5 m straight up. The angle between the action line of the force and the displacement is 0 degrees. Therefore, the work done to the object is:

(98.1 N) * 5 m * cos(0 degrees) = 490.5 joules

Multiple forces on a body

A block on a friction-free surface

A block on a friction-free surface is subjected to three forces:

  1. Weight = 15 N
  2. Upward force from the surface = 15 N
  3. A 5 N force to the right that causes the block to move 5 m to the right at constant velocity.

How much work does each force do on the block.

Answer:

Forces 1 and 2 are perpendicular to the direction of displacement, so they cannot do work on the block.

Force 3 is in the same direction as the displacement and does the following work on the block:

(5 N) * 5 m * cos(0 degrees) = 25 joules

A moving block encounters friction

A block is sliding to the right on a friction free surface. Suddenly the block encounters an area on the surface where the friction is not zero. The friction causes the block to stop after moving 5 m to the right.

When the block encounters the frictional surface, it is subjected to the following three forces:

  1. Weight = 15 N
  2. Upward force from the surface = 15 N
  3. A 5 N force to the left caused by the friction.

How much work does each force do on the block?

Answer:

Forces 1 and 2 are perpendicular to the direction of displacement, so they cannot do work on the block.

The angle between force 3 and the direction of displacement is 180 degrees. Force 3 does the following work on the block:

(5 N) * 5 m * cos(180 degrees) = -25 joules

Note that in this case, the work done is negative.

Another friction scenario

A block on a frictional surface is subjected to four forces:

  1. Weight = 15 N
  2. Upward force from the surface = 15 N
  3. A 5 N force to the right causes the block to move 5 m to the right at constant speed.
  4. A 5 N force to the left caused by friction.

How much work does each force do on the block?

Answer:

Forces 1 and 2 are perpendicular to the direction of displacement, so they cannot do work on the block.

The work done by force 3 is:

(5 N) * 5 m * cos(0 degrees) = 25 joules

The work done by force 4 is:

(5 N) * 5 m * cos(180 degrees) = -25 joules

Force parallel to a ramp

A man and a crate

A man needs to load a crate with friction-free wheels and a weight of 1000 N onto the bed of a truck that is 1 m off the ground. He has two different ramps that he can use to push the crate up the ramp and onto the truck.

One ramp is 2.5 m long and the other ramp is 5 m long.

He is able to push the crate up the ramp by pushing in a direction that is parallel to the ramp.

Questions

What is the force that he must exert parallel to each ramp to push the crate up the ramp.

Which ramp requires the least amount of work to push the crate up the ramp?

Required force

Trigonometry can be used to compute the amount of force required, parallel to the ramp, to push the crate up the ramp as a function of the angle that the ramp makes with the horizontal. That force is equal to the product of the weight of the crate (1000 N or 225 pound-force) and the sine of the angle that the ramp makes relative to the horizontal.

For example, other than the requirement to overcome inertia to get the crate moving, no force is required to push the crate if the angle is 0 degrees. At the other extreme, if the angle is 90 degrees, the man must lift the crate straight up, bearing the entire weight of the crate. For angles in between these two extremes, the required force is within the capability of one man to achieve.

Answers

Ramp #1

  • Truck bed height = 1.0m
  • Ramp length = 2.5m
  • Ramp angle = arcsin(1 / 2.5) = 23.58 degrees
  • Weight of crate = 1000N
  • Component of crate's weight parallel to ramp = 1000 N * sin(23.58 degrees) = 400.03 newtons. This is the force that must be exerted to push the crate up the ramp.
  • Work = (400.03 newtons) * 2.5 meters = 1000.08 joules

Ramp #2

  • Truck bed height = 1.0m
  • Ramp length = 5m
  • Ramp angle = arcsin(1 / 5) = 11.54 degrees
  • Weight of crate = 1000N
  • Component of crate's weight parallel to ramp = 1000 N * sin(11.54 degrees) = 200.05 newtons. This is the force that must be exerted to push the crate up the ramp.
  • Work = (200.05 newtons) * 5 meters = 1000.25 joules

The effect of the ramp

The man only has to exert half as much force to push the crate up Ramp #2 as is required for Ramp #1.

However, he has to push it twice as far with Ramp #2, so the amount of work done on the crate in both cases is 1000 joules.

Like many other simple machines, the use of a ramp (inclined plane) reduces the force required to do a job but it doesn't reduce the amount of work required to do the job.

Do the calculations

I encourage you to repeat the calculations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modules in this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

Note:

Housekeeping material
  • Module name: Energy -- Work for Blind Students
  • File: Phy1170.htm
  • Keywords:
    • physics
    • accessible
    • accessibility
    • blind
    • graph board
    • protractor
    • screen reader
    • refreshable Braille display
    • JavaScript
    • trigonometry
    • work
    • energy
    • power

Note:

Disclaimers:

Financial : Although the Connexions site makes it possible for you to download a PDF file for this module at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should be aware that some of the HTML elements in this module may not translate well into PDF.

I also want you to know that I receive no financial compensation from the Connexions website even if you purchase the PDF version of the module.

Affiliation : I am a professor of Computer Information Technology at Austin Community College in Austin, TX.

-end-

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