Work
Whoa!!!
I do!!! 
I just don’t
haveNow
any energy

F
The force, F, pushes the box for a
short distance. This causes the
box to start moving!!!!!!
Work
If a force, F, is used to move an
object a displacement, d, then the
force does work on the object.
W = Fd
F
d
Units for Work
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

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The units for force are Newtons (N) and the units for
displacement are meters (m).
Since Work is equal to force times displacment. We could
use a Newton-meter (N·m).
But the more common unit is the Joule (J)
1 J = 1 N·m
1 J is defined as the amount of work for 1- N of force to
move an object 1-m
In order for work to be done:
There must be a force applied.
The object must have a displacement.
The force must be responsible for the displacement.
How much work is being done by this weightlifter?
NONE!!!!!
Since he doesn’t move the
weights any distance (There
is no displacement), the
work done is 0 J.
How much work is being done in these cases?
NONE!!!!!
In each case the forces are perpendicular
to the displacement, therefore they are
not “responsible” for the motion.
In general, when a force is perpendicular
to the motion of an object, it DOES NO
WORK!!!!
The weight lifter in the video lifts 263.5-kg a distance of 2.43-m.
How much work does he do during one lift?
If he lifts it twice, how much work would he do?
Watch Video
If more than one force is acting on an object, you can find the total work done on
the object by adding the work done by each force together.
If the force is in the opposite direction as the motion, then the work done is
negative because it removes energy from the object.
If the force is in the same direction as the motion, then the work done is positive
because it adds energy to the object.
F
f
d
A box is being pushed with a force of 25-N. There is also friction
between the box and the table of 12-N. The box is moved 3.0-m
25-N
12-N
3.0-m
How much work is being done by the 25-N force?
F = 25-N
d = 3.0-m
W25 = Fd = (25)(3.0) = 75-J
This work is positive because the force is in the direction of the motion and adds
energy to the box.
How much work is being done by the friction?
f = 12-N
d = 3.0-m
W12 = -fd = -(12)(3.0) = -36-J
This work is negative because the force is in the opposite direction of the motion
and removes energy from the box.
How much total work is being done on the box?
SAME PROBLEM, DIFFERENT METHOD
A box is being pushed with a force of 25-N. There is also friction
between the box and the table of 12-N. The box is moved 3.0-m.
Find the total work done on the box.
25-N
Another way to find the total work done on the
12-N
box is to find the net force on the box and
3.0-m
multiply this by the displacement.
What is the net force on the box?
F = 25-N
f = 12-N
Fnet = F – f = 25 – 12 = 13-N
How much total work is being done on the box?
Fnet = 13-N
d = 3.0-m