When we are told that a person pushes on an object with
a certain force, we only know how hard the person pushes: we don’t
know what the pushing accomplishes. Work, W,
a scalar quantity that measures the product of the force exerted
on an object and the resulting displacement of that object, is a
measure of what an applied force accomplishes. The harder you push
an object, and the farther that object travels, the more work you
have done. In general, we say that work is done by a
force, or by the object or person exerting the
force, on the object on which the force is acting.
Most simply, work is the product of force times displacement. However,
as you may have remarked, both force and displacement are vector
quantities, and so the direction of these vectors comes into play
when calculating the work done by a given force. Work is measured
in units of joules (J), where 1 J = 1 N
· m = 1 kg · m2/s2.
Work When Force and Displacement Are Parallel
When the force exerted on an object is in the same direction
as the displacement of the object, calculating work is a simple
matter of multiplication. Suppose you exert a force of 10
on a box in the northward direction, and the box moves 5
to the north. The work you have done on the box is
N · m = 50
J. If force and
displacement are parallel to one another, then the work done by
a force is simply the product of the magnitude of the force and
the magnitude of the displacement.
Work When Force and Displacement Are Not Parallel
Unfortunately, matters aren’t quite as simple
as scalar multiplication when the force and displacement vectors
aren’t parallel. In such a case, we define work as the product of
the displacement of a body and the component of the force in the
direction of that displacement. For instance, suppose you push a
box with a force F
the floor for a distance s
but rather than pushing it directly forward, you push on it at a
downward angle of 45º
. The work you do on the box is
not equal to
, the magnitude of the force times the magnitude
of the displacement. Rather, it is equal to
, the magnitude of the force exerted in
the direction of the displacement times the magnitude of the displacement.
Some simple trigonometry shows us that
is the angle between
vector and the s
With this in mind, we can express a general formula for the work done
by a force, which applies to all cases:
This formula also applies to the cases where F
parallel, since in those cases,
, so W = Fs
What the formula above amounts to is that work is the
dot product of the force vector and the displacement vector. As
we recall, the dot product of two vectors is the product of the magnitudes
of the two vectors multiplied by the cosine of the angle between
the two vectors. So the most general vector definition of work is:
The concept of work is actually quite straightforward,
as you’ll see with a little practice. You just need to bear a few
simple points in mind:
- If force and displacement are both in the
same direction, the work done is the product of the magnitudes of
force and displacement.
- If force and displacement are at an angle to one another,
you need to calculate the component of the force that points in
the direction of the displacement, or the component of the displacement
that points in the direction of the force. The work done is the
product of the one vector and the component of the other vector.
- If force and displacement are perpendicular, no work is
Because of the way work is defined in physics,
there are a number of cases that go against our everyday intuition.
Work is not done whenever a force is exerted, and there are certain
cases in which we might think that a great deal of work is being
done, but in fact no work is done at all. Let’s look at some examples
that might be tested on SAT II Physics:
- You do work on a 10 kg mass
when you lift it off the ground, but you do no work to hold the
same mass stationary in the air. As you strain to hold the mass in
the air, you are actually making sure that it is not displaced.
Consequently, the work you do to hold it is zero.
- Displacement is a vector quantity that is not the same
thing as distance traveled. For instance, if a weightlifter raises
a dumbbell 1 m, then lowers it to its original position, the weightlifter
has not done any work on the dumbell.
- When a force is perpendicular to the direction of an object’s
motion, this force does no work on the object. For example, say
you swing a tethered ball in a circle overhead, as in the diagram
below. The tension force, T,
is always perpendicular to the velocity, v,
of the ball, and so the rope does no work on the ball.
||A water balloon of mass m is
dropped from a height h. What is the work done
on the balloon by gravity? How much work is done by gravity if the
balloon is thrown horizontally from a height h with
an initial velocity of ?
What is the work done on the balloon by gravity?
Since the gravitational force of –mg is
in the same direction as the water balloon’s displacement, –h,
the work done by the gravitational force on the ball is the force
times the displacement, or W = mgh,
where g = –9.8 m/s2.
How much work is done by gravity if the balloon
is thrown horizontally from a height h with an initial
velocity of v0?
The gravitational force exerted on the balloon is still –mg,
but the displacement is different. The balloon has a displacement
of –h in
the y direction and d (see
the figure below) in the x direction. But, as we
recall, the work done on the balloon by gravity is not simply the product
of the magnitudes of the force and the displacement. We have to
multiply the force by the component of the displacement that is
parallel to the force. The force is directed downward, and the component
of the displacement that is directed downward is –h.
As a result, we find that the work done by gravity is mgh,
just as before.
The work done by the force of gravity is the same if the
object falls straight down or if it makes a wide parabola and lands 100 m
to the east. This is because the force of gravity does no work when
an object is transported horizontally, because the force of gravity
is perpendicular to the horizontal component of displacement.
Work Problems with Graphs
There’s a good chance SAT II Physics may test your understanding
of work by asking you to interpret a graph. This graph will most
likely be a force vs. position graph, though there’s a chance it
may be a graph of
vs. position. Don’t let
the appearance of trigonometry scare you: the principle of reading
graphs is the same in both cases. In the latter case, you’ll be
dealing with a graphic representation of a force that isn’t acting
parallel to the displacement, but the graph will have already taken
this into account. Bottom line: all graphs dealing with work will
operate according to the same easy principles. The most important
thing that you need to remember about these graphs is:
The work done in a force vs. displacement graph
is equal to the area between the graph and the x-axis
during the same interval.
If you recall your kinematics graphs, this is exactly
what you would do to read velocity on an acceleration vs. time graph,
or displacement on a velocity vs. time graph. In fact, whenever
you want a quantity that is the product of the quantity measured
by the y-axis and the quantity measured by the x-axis,
you can simply calculate the area between the graph and the x-axis.
graph above plots the force exerted on a box against the displacement
of the box. What is the work done by the force in moving the box
from x = 2 to x = 4?
The work done on the box is equal to the area of the shaded
region in the figure above, or the area of a rectangle of width 2 and
height 4 plus the area of a right triangle of base 2 and height 2. Determining
the amount of work done is simply a matter of calculating the area
of the rectangle and the area of the triangle, and adding these
two areas together:
Curved Force vs. Position Graphs
If SAT II Physics throws you a curved force vs.
position graph, don’t panic. You won’t be asked to calculate the
work done, because you can’t do that without using calculus. Most
likely, you’ll be asked to estimate the area beneath the curve for
two intervals, and to select the interval in which the most, or
least, work was done. In the figure below, more work was done between x =
6 and x = 8 than between x =
2 and x = 4, because the area
between the graph and the x-axis is larger for
the interval between x = 6 and x = 8.