FFR - Work

Original textbook reading.

Throughout this chapter, we have covered a variety of ways you can encounter forces, and how to deal with them. One topic that is closely related to force is called work. This is not the type of work that earns people money, but instead a way of conveying what a force actually does.

Working with Work

When discussing work, we will be using the SI unit Joule. That might be familiar to you, as it is a unit of energy, a topic we’ll be covering in detail in chapter 5. In this section, you will simply learn about mechanical work, or the energy done by a force on an object.

To start out, we must first figure out how work is calculated.

$W = \vec{F} • \vec{d}$ (Equation 2.4.1)

For a constant force, work equals the dot product of force and displacement, so it is a scalar. The equivalent formula below uses the angle between those vectors.

$W = | \begin{matrix} \vec{F} \end{matrix} | | \begin{matrix} \vec{d} \end{matrix} | \cos θ$ (Equation 2.4.2)
This equation will most likely be used the most, as it is the most useful version unless the vectors are given by components.

One thing that is important to note is that due to the $\cos θ$ being a part of the equation, the force must have some component in the direction of the displacement. In other words, if a force acts perpendicularly to displacement, then it does no work on the object. In a physical sense, you can imagine sliding a book across a table. We know that the force of gravity is acting on the book, as it has weight. However, that force is straight down, while the book’s displacement is across the table. This means that the $θ$ between the two forces is $90^{\circ}$ , making the $\cos θ = 0$ . Therefore, the force of gravity exerts no work throughout the movement of the book.

Another aspect of work that is important to grasp is the concept of negative work. If the force exerted is parallel but opposite to the direction of motion, negative work is done.

For example, take the case of a book on a table. If we were to lift the book straight up, then the force of gravity on the book would be pulling in an opposite direction on the book, resulting in it doing negative work on the object. When we say ‘negative work’ what is really happening is that that force is resisting the motion of the option.In the lifted book example, the force doing the lifting (promoting the motion) exerts a positive work on the book, while the force of gravity (hindering the motion) results in negative work being done on the object.

Example Problem 2.4.1

If a $10 kg$ block is pulled up a ramp by a rope with a force of $100 N$ , and the block moves a distance of $5 m$ up the ramp at an angle of $30^{\circ}$ , give the works done be all the involved forces of the problem. The static of kinetic friction between the ramp and the box is $0.3$ , and it can be assumed that the box will move up the ramp.

Step 1: Drawing the free-body diagram

The forces on the block are gravity, normal force, kinetic friction and the rope’s applied force. Compute the work done by each.

Step 2: Calculate unknown forces.

We start with the normal force, as that is part of the frictional force. The only force in the system with a component in line with the normal force is gravity.

We separate out the force of gravity into its components, figuring out that the normal force has a magnitude of $85.0 N$ .

Using this value for the normal force, we discover that ${\vec{F}}_{f r} = 25.5 N$ .

Step 3: Identify the angles between each involved force and the displacement

By investigating the FBD we drew in step 1, we can come up the angles of all the involved forces. We notice the angle of the applied force from the rope is $0^{\circ}$ .

$θ_{{\vec{F}}_{f r}} = 180^{\circ}$
$θ_{{\vec{F}}_{g}} = 120^{\circ}$
$θ_{{\vec{F}}_{N}} = 90^{\circ}$

Step 4: Calculate the works done by each force

W_g = −245.25 J; W_friction ≈ −127.44 J; W_applied = +500 J; W_normal = 0 J.

Pause & recall

Pop Quiz 2.4.1 Work

Question 1 of 5 · 0 correct

AP Topic 3.2 · Read the worked solution

A box weighs 500 N. Compare the applied work needed to raise it steadily by 5 m vertically with the work needed to push it steadily up a frictionless 30° ramp to the same height. The push is parallel to the ramp. How much less work is needed using the ramp?

Read every full question without answers

Question 1

$2250 J$
0 J
$2211 J$
$2500 J$

Separate worked answer

Question 2

If a $57 g$ tennis ball falls $10 m$ , with air providing a resistive force of $0.25 N$ , what is the net work exerted on the ball?

Question diagram; use the accompanying question and labels.

$6.84 J$
$5.59 J$
$4.34 J$
$3.09 J$

Separate worked answer

Question 3

What force exerts more work, a $20 N$ force exerted at $45^{\circ}$ from the displacement, or a $15 N$ force exerted in line with the displacement? Assume both forces are exerted on the same object, so the distance traveled is equal.

The $15 N$ Force
The $20 N$ Force
They are equal
More information Needed

Separate worked answer

Question 4

If a man pulls on a block with a rope that is at a $60^{\circ}$ angle to the block as it slides along the ground, how much work does the man do in sliding the block $20 m$ ? The man is pulling with a $100 N$ force.

$2000 J$
$1500 J$
$1000 J$
$200 J$

Separate worked answer

Question 5

If a block has 4 forces acting on it, as shown, and if moves horizontally 10m, rank the forces in order of increasing work done. The forces are: $F_{1} = 10 N$ , $F_{2} = 15 N$ , $F_{3} = 10 N$ , $F_{4} = 5 N$ .

1, 4, 3, 2
4, 1, 2, 3
4, 1, 3, 2
1, 4, 2, 3

Separate worked answer

This idea of work is a very useful concept for explaining how much actual energy is needed to move something a certain distance, assuming the work is 100% efficient. Why wouldn’t it be, you might ask? We are about to find out .

Conservatives vs. Non-Conservative Forces

Inspect how each force points relative to the instantaneous displacement.

A force is conservative when its work between any two points is independent of the path, equivalently when its work around every closed path is zero. Work itself is defined for both conservative and nonconservative forces.

Conservative forces also have the fun little identity that no matter what the path an object takes, if the displacement is $0 m$ , then that conservative force will have exerted no work on the object.

A nonconservative force can do nonzero work along a closed path even though the net displacement is zero.

Kinetic friction on a block sliding over a stationary surface does negative work because it opposes each displacement. On a trip up a ramp and back, both legs contribute negative friction work. The dot-product and integral definitions apply on either a straight or curved path.

Example Problem 2.4.2

A 50 kg box moves 5 m uphill on an 18° ramp under a 200 N winch pull parallel to the ramp, then returns 5 m downhill with no winch force. Use μₖ = 0.2 and g = 9.81 m/s². Compute work over this prescribed pair of path segments. The stated pull cannot start this uphill motion from rest; a real rope-break history may include extra coasting not included here.

Step 1: Free Body Diagram

First things first, we start out with an FBD, just like normal. In this case we have to draw 2 FBDs, one for the box as it is being pulled up the ramp, and another for the box after it starts to fall.

Step 2: Find the values for the unknowns

In each FBD, the only unknowns are the normal force and friction. We start with the normal force. In this case, the normal force is only resisting gravity. Therefore, we need the component of gravity in the direction of the normal force. This turns out to be $466.5 N$ .

This makes friction $93.3 N$ .

Step 3: Calculate the work done one the box for the first half of the time

During this first situation, the winch is pulling the box up the ramp for $5 m$ .

Again, we need to find the angles between each force and the displacement. They are:
$\begin{aligned} θ_{{\vec{F}}_{g}} & = 72^{\circ} \\ θ_{{\vec{F}}_{f r}} & = 180^{\circ} \\ θ_{{\vec{F}}_{N}} & = 90^{\circ} \end{aligned}$

This makes the works during this leg of the box’s journey:

On the 5 m downhill segment: W_g = +757.86 J; W_friction = −466.49 J; W_normal = 0 J.

Step 5: Combine each segment of the entire displacement

To find the total work contributions, we merely need to add together each force’s work exertion during each part. This gives us:

For the prescribed round trip: W_g = 0 J, W_friction ≈ −932.99 J, W_winch = +1000 J and W_normal = 0 J. Thus net work is approximately +67.01 J.

To figure out the work done by each force, we can just use the usual approach.

The angles between each force and the displacement are:

On the 5 m uphill segment: W_g = −mg(5 sin 18°) = −757.86 J; W_friction = −μₖmg cos 18°(5) = −466.49 J; W_winch = +1000 J; W_normal = 0 J.

Step 4: Calculate the work done over the second half

For this, the winch force is not active, so the only forces present are the normal force, friction, and gravity.

Pause & recall