Chapter 7: Work & Kinetic Energy

7.0. Overview

We've had an introduction to Newton's Laws of Motion, and how we can use free-body diagrams with Fnet = ma to solve some interesting problems. In this section we'll continue our studies, using the concept of Force applied to a mass while it travels some displacement (distance in a direction).

Let's get started!

  1. Definition of Work
  2. The dot product
  3. Work done by a varying Force
  4. The work-energy theorem
  5. Definition of Power

7.1. Definition of Work

Work is a quantity that is useful in describing how objects interact with other objects.

Definition of Work

Work done by an agent exerting a constant force on an object is the product of the component of the force in the direction of displacement, and the magnitude of the displacement of the object.

Note that the word "work" is used in a lot of ways that don't correspond with "physics work":

"Physics work" occurs when the conditions described above are met: when a Force is applied to an object in a direction that is moving over some displacement. Let's see some examples of that.

Work done (frictionless)

A constant 100-N horizontal force is used to pull a 20kg box 2.0-m across a frictionless table. How much Work is done on the box in each of the following situations?

  1. How much Work is done by the horizontal Force?

    2. Solution:

    Begin, of course, by drawing a free-body diagram of the situation, because we're looking at Work done by specific force, and we therefore need to have identified that force. Not all work-energy problems focus on forces, but as we're developing these ideas, we'll want to do a good job demonstrating the process.

    The simplest way to solve this problem is to use the "Work equals force times distance" formula, where the force-applied of 100 N and the distance traveled along that line of force is 2.0 meters.

    Using this formula assumes that the Force and the distance traveled are in the same direction.

    It's important to know that the Work formula is actually based on vectors, so the direction of the Force and the direction of the displacement of travel are what we're really using. In this case, those two vectors are in the same direction, ie. there is an angle θ of 0 degrees between them. Applying the more formal analysis yields the same result:

  2. How much Work is done by the force down due to gravity?

    3. Solution:

    Although the force of earth's gravity pulls down on the box with a force of mg = 196 Newtons, there is no displacement in that direction. Therefore, the force of gravity does no work on the box.

    We can demonstrate this more formally by writing:

  3. How much Work is done by the normal force up from the table?

    4. Solution:

    Using the same reasoning as above, we can determine that the Work done by the Normal force from the table is 0 Joules.

Work done (friction)

The same constant 100-N horizontal force is used to pull a 20-kg box 2.0-m across a rough surface at constant velocity.

How much Work is done by the applied force now? How much Work is done by the force of friction?

Solution:

If the force applied and displacement remain the same, then the work done by the force applied remains the same: 200J.

To calculate the Work done by friction, consider the force of friction acting in the 180 degree direction and the displacement of the block along the table, in the positive-x, 0 degree direction.

The θ between those two vectors is 180 degrees, and that's what we use to calculate the Work done by friction, then:

Work is not a vector quantity

It's important to realize that, although Work is calculated using vector quantities, it is itself a scalar quantity: it doesn't have a direction.

Forces acting on a moving object may "do positive work on the object," in which case (we'll soon see) the force is increasing the energy of the object. A force may also "do negative work on the object," in which case the force is decreasing the energy of the object.

The positive and negative Work values calculated refer to energy added to or removed from the object, not "the direction of the work."

Work is a scalar quantity that doesn't have a direction.

Work done (frictionless), force at an angle

A constant 100-N is applied at 30° above the horizontal, and used to drag the 20-kg box 2.0-m across a frictionless surface.

How much work is done by the applied force?

Solution:

There are a couple of ways you can think about this problem. One way is to look at the displacement in the positive-x direction, and make sure that you only use the component of the Force applied that aligns with that displacement. That's Fx, which we can see from the diagram is F cos θ. In that case:

Using the other definition of work, with θ indicating the angle between the two vectors, leads to the same result:

Work done by friction, force at an angle

A 100-N is applied at 30° above the horizontal, and used to drag the 20-kg box 2.0-m across a rough surface (μ = 0.15).

How much work is done by friction?

Solution:

Again, the force of friction and the displacement are in opposite directions, so we can expect that the work done by the force of friction will be negative. But how do we actually calculate the force of friction?

If we knew that the box was being pulled at a constant velocity we'd know that the Fapplied and the Ffriction were the same. But we haven't been given that information. We can calculate Friction using μ and the normal force, though.

Just be sure to calculate the correct Normal force. With the force applied pulling up on the box, the normal force is no longer equal to mg!

Does friction always do negative work?

No. In some circumstances friction does positive work.

Consider the case of a waiter carrying a serving tray with a drink on it. As the serving tray is accelerated sideways (to the right in the diagram shown here), there is a force of static friction between the tray and the glass. The force of friction between those two objects is shown as force-pairs: the force on the drink from the tray, and the third-law-pair, the force on the tray from the drink. The drink's inertia is such that it "wants" to remain at rest, but the friction force from the tray (to the right) and the displacement of the drink (to the right) are in the same direction.

In this situation, the force of friction is doing positive work, Ffriction x.

Drawing good free-body diagrams is essential to analyzing these problems!

The apple

  1. I lift an 0.306-kg (3.00 Newtons) apple up 50-cm into the air at constant velocity—how much Work did I do?
  2. How much Work did earth's gravity do in the preceding problem?
  3. How much Work do I do lowering a 3.00-N apple 50-cm down?
  4. How much Work do I do holding a 3.00-N apple motionless in the air?
  5. How much Work do I do carrying a 3.00-N apple sideways 50-cm at constant velocity?

Solution:

  1. The apple is moving at constant speed upwards, so the net force on it is 0. Therefore the force I apply is equal to the force of gravity, or 3.0-N. The force I apply and the displacement are in the same direction, so:
  2. The force of earth's gravity is pulling in the opposite direction of the displacement, so we would expect the Work to be negative, and so it is:
  3. This is an interesting problem. I'm lowering the apple at constant velocity, which involves balanced forces—earth's gravity pulling down with 3.0 N and me pushing up with 3.0 N—as the ball moves downwards. Because I'm lifting up as the apple is moving down, the force and displacement are in the opposite direction. Based on previous calculations, I am doing -1.5 J of work on the apple.
  4. Although I'm applying a force of 3.0 N upwards, if there is no displacement in the direction of Force, there is no work being done.
  5. This is really interesting: there is a Force upwards from my hand supporting the apple, but the apple has a sideways displacement. The Work done by me under these conditions is:

    A great question to ask would be "Wait, aren't you applying a little bit of force to move the apple sideways?" And it's true, if the apple weren't moving to begin with and we wanted to get it going sideways, we'd have to apply a net force sideways to start it moving. During that brief period of time, we are doing Work on the apple, but once it's moving sideways, its inertia carries it along. The only force we have to continue to apply is upwards against gravity, and that upwards force, as we've determined, is doing no Work on the apple.

7.2. The dot product

The concept of a vector occurs in physics, of course, but it exists in mathematics, as well as in computer science. The application of vectors in each field varies a little, but the principles are the same.

We are going to extend our definition of Work to include a mathematical definition of one kind of vector multiplication, called the "dot product."

7.2.1. Definition of the dot product

We've already seen how a scalar quantity multiplied by a vector produces a new vector pointing in the same direction, but with a magnitude changed by the amount of the scalar. 3i, for example, is the unit vector i, but with a magnitude of 3.

If instead of a scalar and a vector you have two vectors, there are two useful ways to multiply them together, and we're going to use both of them in this course. The first way you can multiply two vectors is using the "dot product":

The "dot product," indicated appropriately enough with a "dot," is a measure of how parallel two vectors are, and is easily calculated as follows (using two-dimensional x-y vectors as an example):

It can be shown (although we don't do it here) that this dot-product is equal to ABcosθ... which is the formula that we use to calculate Work! Therefore:

If you need to calculate Work done, and F and x have been given to you in i, j notation, you can use the dot-product to easily solve that problem.

Dot-product example

If vector A = 3i + 5j, and vector B = -1i + 2j:

  1. Draw a sketch of the two vectors, each one starting at the origin, (0, 0).
  2. What is the dot-product A • B ?
  3. What is the angle between the vectors A and B ?

Solution:

Dot-product Work example

A force of 4i + -3j Newtons acts on object, and displaces it 10i + 3j meters. How much Work was done no the object by the force?

Solution:

For a theoretical problem like this, it's simple enough to apply the dot-product process to calculate the Work done.

Remember that the dot-product produces a scalar result—there is no direction associated with the result, even if we are multiplying vectors in the process.

7.3. Work done by a varying Force

Up to this point we've looked at the work done by a constant force, a Force applied or the force due to earth's gravity. There are a wide variety of situations in which a force varies as a function of position: a rubber band, for example, pulls back harder the more you stretch it out.

If you're trying to solve for Work done by "an average force of some-number-of Newtons...", the simple equation works fine. If the Force varies as a function of position, you'll need to do some calculus.

The small amount of Work done as a Force is applied over a small displacement:

The total amount of Work done by a Force that varies as a function of displacement:

7.3.1. Two ways to determine work done by a varying Force

If the force is varying, we've got two common strategies for determining how much Work is being done over some displacement:

  1. Use a graph of Force vs. displacement to identify the area under the curve

    Here, the width of each column is 1 meter, and the amount of Work done by the Force during that displacement is simply the height of the column (F) times the width of the column (1 meter). The total Work done is found by identifying the total area under the curve.

  2. Use a given F(x) function and integrate with respect to x.

    If you've got a function for force F as a function of position, you can perform an integration over the interval xinitial to xfinal to determine the total amount of Work done.

Let's see how to use each of these strategies.

Area Under the Curve

Using the graph shown above, determine the total amount of Work done by the force while moving a mass from 0 to 10 m along the x-axis.

Solution:

Add up the squares for the area-under-the-curve from 0 to 10 m to get a total Work of 22.5 Joules.

7.3.2. Hooke's Law

As mentioned above, a rubber band or a spring is a good example of a device that applies a varying Force depending on how far it has been stretched out. An ideal spring is one that follows "Hooke's Law."

Hooke's Law

Hooke's Law describes the behavior of some springs under a limited range of conditions, in which the force applied by the spring (either stretched or compressed) is linearly proportional, and in the opposite direction of, its displacement.

The value k is called the "spring constant," and describes the Force-per-meter necessary to compress or stretch the spring.

The negative in the equation indicates that the Force the spring applies is in the direction opposite of the displacement x. The spring constant k is always a positive value.

Calculate the spring constant

A 100-gram mass is attached to one end of a hanging spring and slowly released so that it comes to rest, stretching out the spring 50-centimeters. Calculate the spring constant of the spring.

 

Solution:

Once the mass has been attached and placed in a position where it is at rest, the force of earth's gravity pulling down on the mass and the force of the spring pulling up on it are equal to each other. Knowing the Force that the spring is applying and the extension x of the spring we can calculate:

7.3.3. Integrating F(x) with a mass-spring system

As we develop some of these ideas we'll use this situation as our context: a mass, resting on a smooth (frictionless) horizontal surface, with one end of the spring attached to an anchor at one end.

Definitions:

Work done by a spring as it expands

A block with mass m is attached to a spring of spring constant k and arranged horizontally on a frictionless surface with the spring anchored to one end. The spring is compressed from its equilibrium position by a distance x and then released. How much Work is done by the spring to return the block to its equilibrium position?

Solution:

This is an important solution that you should make sure you have a good understanding of before moving on.

We'll take the definition of the Work integral, and substitute in -kx from Hooke's Law, and then evaluate the integral, integrating with respect to x.

This is the amount of Work that the spring does in returning the mass to its equilibrium position.

Note that the mass doesn't stop once it returns to the equilibrium position in this situation! Its inertia carries it past equilibrium, and now that the displacement is in the positive direction, the force from the spring begins to pull it back in the opposite direction, slowing it down.

We'll discuss this kind of oscillating motion later on in the course.

Now how much Work?

Continuing the problem above, if we were to compress the spring twice as much (2x), how much Work would the spring do (compared with the previous answer) in returning the mass to the equilibrium position?

Solution:

Given that x2 term in the previous result, we might expect the Work done to increase by a factor of 22, to give us "four times the Work" as an answer. Solve the problem again with limits of 0 and 2x to see if that's correct. :)

7.4. The work-kinetic energy theorem

If force F varies in a problem, we can't use our constant-acceleration kinematics to analyze the motion... but we can use the concept of Work.

Take a look at this important derivation that uses two important equations—Newton's Second Law of Motion and the Work integral—to come up with a new relationship.

Derivation of the work-kinetic energy theorem

Let's begin with the definition of Work and Newton's 2nd Law:

We've got a little bit of an issue here: m is constant, but acceleration a is changing, so that's what we need to integrate... but we can't integrate a with respect to x. Let's use the Chain Rule to convert a to something else:

Substituting back into the integral...

Here we have a couple of dx expressions. We're integrating with respect to x, and also differentiating with respect to x. Those two operations are inverses of each other, so they effectively cancel each other out.

Because we are now integrating with respect to velocity v, we have to change our limits. At position xi we have velocity vi, and at the final x we had a final v. Finally, evaluate the integral.

This equation reveals that a force doing Work on a moving mass will change the mass's velocity.

7.4.1. Definition of kinetic energy

You may recognize the expression . That expression appears so frequently in our calculations that it has it's own "shortcut" variable, K.

Definition of kinetic energy

The expression is referred to as "the kinetic energy of a mass," and is represented by the variable K.

The three bars in the "equals" sign indicate that the expression "K" is defined as . These two expressions are equivalent to each other.

Thus, Work that causes a change in kinetic energy can be written in a few different ways:

Recall that Work done can be positive or negative. Positive work has the effect of increasing the kinetic energy of an object, while negative work decreases the kinetic energy of the object.

7.4.2. Solving some Work-energy problems

Accelerating a block

A 6.00 kg mass starting at rest is pulled with a constant horizontal force of 12.0 N for a distance of 3.00 m, accelerating the mass across a frictionless surface. Find the final speed of the block using two different strategies.

  1. Strategy 1: Calculate the acceleration of the mass, and then use kinematics to find the final speed of the block after it has traveled the 3.00 m.

    2. Solution:

  2. Strategy 2: Calculate the Work done on the block by the mass, and then use the work-energy theorem to find the final speed of the block.

    3. Solution:

    Which strategy "feels" better to you in solving this problem? Which strategy is more adaptable to a variety of situations?

Work done by a spring

A dart is loaded into a spring-driven Nerf gun by pushing the spring in for a distance d. For the next loading, the spring is compressed a distance 2d. How much faster does the second dart leave the gun, compared to the first?

Solution:

Strategy: find the Work that the spring does on the dart to change its kinetic energy.

So, twice the compression x implies that the velocity as the dart leaves the gun will be twice the original speed.

7.5. Definition of Power

Power in physics refers to the time-rate at which Work is done. If a force is doing some Work, then the more quickly the work is done, the more Power is necessary to do that Work.

Definition of Power

If a force is doing Work over some period of time we can calculate the average Power necessary to do that work. If we have a time-based function for the Work being done, we can also calculate the instantaneous Power being used to do that work.

The units for Power are Watts, where 1 Watt = 1 Joule / 1 second.

An interesting result of this is that we can calculate instantaneous power used by a force if we know the force and the velocity of the moving object.

Lifting a person

At a wedding celebration, the groom sits in a chair and his friends lift up against gravity, using 800 Newtons of force to raise him 2.0 meters into the air. It took them about 4 seconds to lift him up.

  1. How much Work was required to lift the groom-and-chair into the air?
  2. How much average Power was used in the lifting?

Solution:

  1. We can calculate the Work done by the friends as a simple force-times-distance calculation. It's true that the friends have to accelerate the chair when they first start lifting, but then they slow it down at the end as the groom reaches the upper height—thus, the net force exerted by them during the lift is 800 N, and the total Work during the lift is:
  2. To calculate the average Power, use the given time along with the Work just calculated:

Lifting a person (more quickly)

At the same wedding celebration, the same groom sits in the same chair and his friends lift up against gravity, using 800 Newtons of force to raise him 2.0 meters into the air. The friends take only 1 second to lift him up.

  1. How much Work was required to lift the groom-and-chair into the air this time?
  2. How much average Power was used in the lifting?

Solution:

  1. Although the Work is done more quickly this time, the actual amount of Work is the same as it was before:
  2. Because the work has been done four times faster, however, we expect that the Power used will be four times as much, and that turns out to be true:

Lifting an elevator

An elevator with a mass of 1000 kg carries a load of 800 kg. A 4000 N force of friction retards the elevator's upward motion as it moves upwards

  1. Find the minimum power necessary to lift the elevator at a constant speed of 3.00 m/s.
  2. If the motor needs to have a 3:1 safety factor, what should the horsepower rating on the motor be? (746W = 1 hp)
  3. What Power must the motor deliver at any instant (as a function of v ) if its designed to provide an acceleration of 1.00 m/s2?

Solution:

  1. There are two ways to calculate Power, as we've seen. In this case, the more convenient formula is P = Fv.
  2. A three-to-one safety factor means that we need the motor to be three times as powerful as we think we need.
  3. Power as a function of v? Because P = Fv, they're really just asking us to calculate the FTension under these new conditions.