 College Physics for AP® Courses

# 7.1Work: The Scientific Definition

College Physics for AP® Courses7.1 Work: The Scientific Definition

### Learning Objectives

By the end of this section, you will be able to:

• Explain how an object must be displaced for a force on it to do work.
• Explain how relative directions of force and displacement of an object determine whether the work done on the object is positive, negative, or zero.

The information presented in this section supports the following AP® learning objectives and science practices:

• 5.B.5.1 The student is able to design an experiment and analyze data to examine how a force exerted on an object or system does work on the object or system as it moves through a distance. (S.P. 4.2, 5.1)
• 5.B.5.2 The student is able to design an experiment and analyze graphical data in which interpretations of the area under a force-distance curve are needed to determine the work done on or by the object or system. (S.P. 4.5, 5.1)
• 5.B.5.3 The student is able to predict and calculate from graphical data the energy transfer to or work done on an object or system from information about a force exerted on the object or system through a distance. (S.P. 1.5, 2.2, 6.4)

### What It Means to Do Work

The scientific definition of work differs in some ways from its everyday meaning. Certain things we think of as hard work, such as writing an exam or carrying a heavy load on level ground, are not work as defined by a scientist. The scientific definition of work reveals its relationship to energy—whenever work is done, energy is transferred.

For work, in the scientific sense, to be done on an object, a force must be exerted on that object and there must be displacement of that object in the direction of the force.

Formally, the work done on a system by a constant force is defined to be the product of the component of the force in the direction of motion and the distance through which the force acts. For a constant force, this is expressed in equation form as

$W = ∣ F ∣ cos θ ∣ d ∣ , W = ∣ F ∣ cos θ ∣ d ∣ , size 12{W= lline F rline left ("cos"θ right ) lline d rline } {}$
7.1

where $WW size 12{W} {}$ is work, $dd size 12{d} {}$ is the displacement of the system, and $θθ size 12{θ} {}$ is the angle between the force vector $FF size 12{F} {}$ and the displacement vector $dd size 12{d} {}$, as in Figure 7.2. We can also write this as

$W=Fd cosθ.W=Fd cosθ. size 12{W= ital "Fd"" cos"θ} {}$
7.2

To find the work done on a system that undergoes motion that is not one-way or that is in two or three dimensions, we divide the motion into one-way one-dimensional segments and add up the work done over each segment.

### What is Work?

The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts. For one-way motion in one dimension, this is expressed in equation form as

$W=Fdcosθ,W=Fdcosθ, size 12{W= ital "Fd"" cos"θ} {}$
7.3

where $WW size 12{W} {}$ is work, $FF size 12{F} {}$ is the magnitude of the force on the system, $dd size 12{d} {}$ is the magnitude of the displacement of the system, and $θθ size 12{q} {}$ is the angle between the force vector $FF size 12{F} {}$ and the displacement vector $dd size 12{d} {}$.

Figure 7.2 Examples of work. (a) The work done by the force $FF size 12{F} {}$ on this lawn mower is $FdcosθFdcosθ size 12{ ital "Fd""cos"θ} {}$. Note that $FcosθFcosθ size 12{F"cos"θ} {}$ is the component of the force in the direction of motion. (b) A person holding a briefcase does no work on it, because there is no displacement. No energy is transferred to or from the briefcase. (c) The person moving the briefcase horizontally at a constant speed does no work on it, and transfers no energy to it. (d) Work is done on the briefcase by carrying it up stairs at constant speed, because there is necessarily a component of force $FF size 12{F} {}$ in the direction of the motion. Energy is transferred to the briefcase and could in turn be used to do work. (e) When the briefcase is lowered, energy is transferred out of the briefcase and into an electric generator. Here the work done on the briefcase by the generator is negative, removing energy from the briefcase, because $FF size 12{F} {}$ and $dd size 12{d} {}$ are in opposite directions.

To examine what the definition of work means, let us consider the other situations shown in Figure 7.2. The person holding the briefcase in Figure 7.2(b) does no work, for example. Here $d=0d=0 size 12{d=0} {}$, so $W=0W=0 size 12{W=0} {}$. Why is it you get tired just holding a load? The answer is that your muscles are doing work against one another, but they are doing no work on the system of interest (the “briefcase-Earth system”—see Gravitational Potential Energy for more details). There must be displacement for work to be done, and there must be a component of the force in the direction of the motion. For example, the person carrying the briefcase on level ground in Figure 7.2(c) does no work on it, because the force is perpendicular to the motion. That is, $cos90º =0cos90º =0 size 12{"cos""90""°="0} {}$, and so $W=0W=0 size 12{W=0} {}$.

In contrast, when a force exerted on the system has a component in the direction of motion, such as in Figure 7.2(d), work is done—energy is transferred to the briefcase. Finally, in Figure 7.2(e), energy is transferred from the briefcase to a generator. There are two good ways to interpret this energy transfer. One interpretation is that the briefcase’s weight does work on the generator, giving it energy. The other interpretation is that the generator does negative work on the briefcase, thus removing energy from it. The drawing shows the latter, with the force from the generator upward on the briefcase, and the displacement downward. This makes $θ=180ºθ=180º size 12{θ="180"°} {}$, and $cos 180º=–1cos 180º=–1 size 12{"cos 180"°= +- 1} {}$; therefore, $WW size 12{W} {}$ is negative.

### Real World Connections: When Work Happens

Note that work as we define it is not the same as effort. You can push against a concrete wall all you want, but you won’t move it. While the pushing represents effort on your part, the fact that you have not changed the wall’s state in any way indicates that you haven’t done work. If you did somehow push the wall over, this would indicate a change in the wall’s state, and therefore you would have done work.

This can also be shown with Figure 7.2(a): as you push a lawnmower against friction, both you and friction are changing the lawnmower’s state. However, only the component of the force parallel to the movement is changing the lawnmower’s state. The component perpendicular to the motion is trying to push the lawnmower straight into Earth; the lawnmower does not move into Earth, and therefore the lawnmower’s state is not changing in the direction of Earth.

Similarly, in Figure 7.2(c), both your hand and gravity are exerting force on the briefcase. However, they are both acting perpendicular to the direction of motion, hence they are not changing the condition of the briefcase and do no work. However, if the briefcase were dropped, then its displacement would be parallel to the force of gravity, which would do work on it, changing its state (it would fall to the ground).

### Calculating Work

Work and energy have the same units. From the definition of work, we see that those units are force times distance. Thus, in SI units, work and energy are measured in newton-meters. A newton-meter is given the special name joule (J), and $1 J=1 N⋅m=1 kg⋅m2/s21 J=1 N⋅m=1 kg⋅m2/s2 size 12{1" J"=1" N" cdot m=1" kg" cdot m rSup { size 8{2} } "/s" rSup { size 8{2} } } {}$. One joule is not a large amount of energy; it would lift a small 100-gram apple a distance of about 1 meter.

### Example 7.1

#### Calculating the Work You Do to Push a Lawn Mower Across a Large Lawn

How much work is done on the lawn mower by the person in Figure 7.2(a) if he exerts a constant force of $75.0N75.0N size 12{"75" "." 0" N"} {}$ at an angle $35º35º size 12{"35"°} {}$ below the horizontal and pushes the mower $25.0m25.0m size 12{"25" "." 0" m"} {}$ on level ground? Convert the amount of work from joules to kilocalories and compare it with this person’s average daily intake of $10,000kJ10,000kJ size 12{"10","000"" kJ"} {}$ (about $2400kcal2400kcal size 12{"2400"" kcal"} {}$) of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by $1ºC1ºC size 12{1°C} {}$, and is equivalent to $4.184J4.184J size 12{4 "." "184"" J"} {}$, while one food calorie (1 kcal) is equivalent to $4184J4184J size 12{"4184"" J"} {}$.

#### Strategy

We can solve this problem by substituting the given values into the definition of work done on a system, stated in the equation $W=FdcosθW=Fdcosθ size 12{W= ital "Fd"" cos"θ} {}$. The force, angle, and displacement are given, so that only the work $WW size 12{W} {}$ is unknown.

#### Solution

The equation for the work is

$W=Fdcosθ.W=Fdcosθ. size 12{W= ital "Fd"" cos"θ} {}$
7.4

Substituting the known values gives

W = 75.0 N 25.0 m cos 35.0º = 1536 J = 1.54 × 10 3 J. W = 75.0 N 25.0 m cos 35.0º = 1536 J = 1.54 × 10 3 J. alignl { stack { size 12{W= left ("75" "." "0 N" right ) left ("25" "." "0 m" right )"cos " left ("35" "." 0° right )} {} # size 12{" "="1536"" J"=1 "." "54" times "10" rSup { size 8{3} } " J" "." } {} } } {}
7.5

Converting the work in joules to kilocalories yields $W=(1536J)(1kcal/4184J)=0.367kcalW=(1536J)(1kcal/4184J)=0.367kcal size 12{W= $$"1536"J$$ $$1"kcal"/"4184"J$$ =0 "." "367""kcal"} {}$. The ratio of the work done to the daily consumption is

$W 2400 kcal = 1 . 53 × 10 − 4 . W 2400 kcal = 1 . 53 × 10 − 4 . size 12{ { {W} over {"2400"`"kcal"} } =1 "." "53" times "10" rSup { size 8{ - 4} } "." } {}$
7.6

#### Discussion

This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released in the consumption of food is used to do work. Even when we “work” all day long, less than 10% of our food energy intake is used to do work and more than 90% is converted to thermal energy or stored as chemical energy in fat.

### Applying the Science Practices: Boxes on Floors

Plan and design an experiment to determine how much work you do on a box when you are pushing it over different floor surfaces. Make sure your experiment can help you answer the following questions: What happens on different surfaces? What happens if you take different routes across the same surface? Do you get different results with two people pushing on perpendicular surfaces of the box? What if you vary the mass in the box? Remember to think about both your effort in any given instant (a proxy for force exerted) and the total work you do. Also, when planning your experiments, remember that in any given set of trials you should only change one variable.

You should find that you have to exert more effort on surfaces that will create more friction with the box, though you might be surprised by which surfaces the box slides across easily. Longer routes result in your doing more work, even though the box ends up in the same place. Two people pushing on perpendicular sides do less work for their total effort, due to the forces and displacement not being parallel. A more massive box will take more effort to move.

### Applying the Science Practices: Force-Displacement Diagrams

Suppose you are given two carts and a track to run them on, a motion detector, a force sensor, and a computer that can record the data from the two sensors. Plan and design an experiment to measure the work done on one of the carts, and compare your results to the work-energy theorem. Note that the motion detector can measure both displacement and velocity versus time, while the force sensor measures force over time, and the carts have known masses. Recall that the work-energy theorem states that the work done on a system (force over displacement) should equal the change in kinetic energy. In your experimental design, describe and compare two possible ways to calculate the work done.

Sample Response: One possible technique is to set up the motion detector at one end of the track, and have the computer record both displacement and velocity over time. Then attach the force sensor to one of the carts, and use this cart, through the force sensor, to push the second cart toward the motion detector. Calculate the difference between the final and initial kinetic energies (the kinetic energies after and before the push), and compare this to the area of a graph of force versus displacement for the duration of the push. They should be the same.

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