Work and Energy
Written by tutor Tony S.
Work is a physical quantity that is defined in terms of a force causing a displacement of an object. For example, if a person pushes on a box, and that box moves some distance, the person has done work on the box. We can calculate exactly how much work was done by using the following formula:
W = Fd
where W is work, F is force, and d is displacement. The standard SI unit of work is the joule (J), which is equal to the work done by a one-newton force in moving an object one meter:
1 J = 1 N · 1 m
Let’s go back to the previous example. If the person pushes with a constant, ten-newton force, and the box moves a distance of five meters in the direction the person has pushed, then the person has done fifty joules of work on the box:
10.0 N · 5.00 m = 50.0 J
But what happens if the direction the person pushes isn’t parallel to the direction the box moves? Let’s say the box isn’t very high, and the person has to hunch over to push it, causing the direction of his push to be directed at an angle of twenty degrees below the horizontal:
In this case, not all of the force contributes to moving the box. Only the horizontal component of the force vector, F cosΘ, is responsible for the work done on the box. We can incorporate that into our earlier equation for work, giving:
W = Fd cosΘ
Now that the person is hunched over and exerting his force at an angle to the displacement, he is doing less work on the box:
10.0 N · 5.00 m · cos20.0° = 47.0 J
(For those of you familiar with vector calculus, we can represent work as the dot product of the force vector F and the displacement vector d:
W = F · d = Fd cosΘ
where F and d are the magnitudes of the force and displacement vectors, respectively, and Θ is the angle between the two vectors.)
The addition of cosΘ to our formula brings some interesting concepts to light:
- When Θ equals zero, cosΘ equals one. This happens when the force and displacement are in the same direction, and thus we have our original formula, W=Fd.
- When Θ equals 90°, cosΘ equals zero which in turn makes work zero. This happens when the force and displacement are perpendicular to each other. Thus, when a force is applied to an object in a direction that is perpendicular to the object’s displacement, that force does no work.
- When Θ is between 90° and 180°, cosΘ is less than zero which in turn makes work less than zero. This gives us the concept of negative work, which can be illustrated by adding friction to our previous box-pushing scenario.
Let’s say the introduced kinetic friction force is a constant, nine-newton force. We already calculated the work done by the person as forty-seven joules. We can now calculate the work done by friction, realizing that the angle between the displacement and friction vectors is 180°:
9.00 N · 5.00 m · cos180° = -45.0 J
We can add this work to the work done by the person to get the overall net work done on the box, which would be two joules. Instead of finding the work due to each force individually and then summing them, we can calculate the net force first and use that to calculate the net work:
ΣW = (ΣF)d cosΘ
where ΣW is the net work and ΣF is the magnitude of the net force. You can prove this by calculating the net force for the above scenario, then calculating the work. (You should find that the net force is 3.44 N at an angle of 83.4° below the horizontal.)
Energy is another physical quantity, certain types of which are closely related to work. The first type we will discuss is kinetic energy, which is the energy an object has simply by being in motion. We can calculate an object’s kinetic energy with the following formula:
K = 1/2mv2
where K is kinetic energy, m is mass, and v is velocity. (You can use this equation to verify that the units of energy are the same as those of work: joules.)
In our previous example of the man pushing the box, let’s say the box weighs ten kilograms. We can do some calculations to determine what the box’s acceleration is based on the net force it experiences (0.0395 m/s²) and then some more calculations to find out what velocity the box has after it’s been pushed for five meters (0.629 m/s) to finally get us to the point where we can calculate the box’s final kinetic energy:
K = 1/2·10kg·(0.629m/s)2 = 1.98J
That’s a lot of calculations! Thankfully, there’s an easier way to calculate kinetic energy based on net work, and it’s called the work-energy principle:
The net work done on an object is equal to the change in the object’s kinetic energy.
ΣW = ΔK = Kf – Ki = 1/2m(vf2-vi2)
(The f and i subscripts indicate final and initial, respectively.) And, if you remember from the previous example, the net work was calculated to be about two joules. Voilà! Moving forward from this point in your physics studies, you should find that you will use the work-energy principle as much as (if not more than) your previous battery of kinematics equations to calculate how fast an object is moving.
Another type of energy is potential energy, which is defined in the context of certain types of forces we call conservative forces. We won’t go into why certain forces are conservative and others are not at this time. Just know that conservative forces (like gravity and the elastic force) have associated potential energies and nonconservative ones (like friction, air resistance, and the push or pull of a person) don’t.
In general, the change in potential energy associated with a conservative force is the negative of the work done by that force. We can use this information to get formulas for our two types of potential energy:
• Gravitational: Imagine a box weighing five kilograms is lifted ten meters straight up. The work done by gravity in this case is:
Wg = FgdcosΘ = mgdcosΘ = 5kg·9.8m/s·10m·cos180° = -490 J
and thus the change in potential energy is (positive) 490 joules.But what if we moved that same box along a twenty-meter path like so:
Now the work done by gravity is:
Wg = FgdcosΘ = mgdcosΘ = 5kg·9.8m/s·20m·cos120° = -490 J
and the change in potential energy is the same as before. Why is that? Well, if you look at how much the height of the box changes in each of the two scenarios (by using a little trigonometry in the second case), you’ll see that in both cases it’s the same: ten meters. Thus, we can infer that gravitational potential energy depends on the change in height of an object. d cosΘ represents the change in height; we can replace this with h in our formula to get the more common expression for gravitational potential energy:
Ug = mgh
where Ug is gravitational potential energy, m is mass, g is the acceleration due to gravity, and h is the object’s height above some reference point.