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## Work and energy

Current time:0:00Total duration:5:48

# Work and the work-energy principle

AP.PHYS:

INT‑3.E (EU)

, INT‑3.E.1 (EK)

, INT‑3.E.1.1 (LO)

, INT‑3.E.1.4 (LO)

## Video transcript

In order to transfer
energy to an object, you've got to exert a
force on that object. The amount of energy
transferred by a force is called the work
done by that force. The formula to
find the work done by a particular
force on an object is W equals F d cosine theta. W refers to the work done by
the force F. In other words, W is telling you
the amount of energy that the force F is
giving to the object. F refers to the size of
the particular force doing the work. d is the displacement of
the object, how far it moved while the force
was exerted on it. And the theta and
cosine theta refers to the angle between
the force doing the work and the
displacement of the object. You might be wondering what this
cosine theta is doing in here. This cosine theta
is in this formula because the only part of
the force that does work is the component that
lies along the direction of the displacement. The component of the force
that lies perpendicular to the direction of motion
doesn't actually do any work. We notice a few things
about this formula. The units for work are
Newton's times meters, which we called joules. Joules are the same unit
that we measure energy in, which makes sense because
work is telling you the amount of joules
given to or taken away from an object or a system. If the value of the
work done comes out to be positive for
a particular force, it means that that
force is trying to give the object energy. The work done by a
force will be positive if that force or a
component of that force points in the same direction
as the displacement. And if the value of the work
done comes out to be negative, it means that that force is
trying to take away energy from the object. The work done by a
force will be negative if that force or a
component of that force points in the opposite
direction as the displacement. If a force points in
a direction that's perpendicular to
the displacement, the work done by
that force is 0, which means it's neither
giving nor taking away energy from that object. Another way that the work
done by a force could be 0 is if the object doesn't
move, since the displacement would be 0. So the force you exert by
holding a very heavy weight above your head does not
do any work on the weight since the weight is not moving. So this formula
represents the definition of the work done by
a particular force. But what if we wanted to
know the net work or total work done on an object? We could just find
the individual amounts of work done by each particular
force and add them up. But there's actually a
trick to figuring out the net work done on an object. To keep things simple, let's
assume that all the forces already lie along the
direction of the displacement. That way we can get rid
of the cosine theta term. Since we're talking about the
net work done on an object, I'm going to replace F with
the net force on that object. Now, we know that the
net force is always equal to the mass
times the acceleration. So we replace F
net with m times a. So we find that the net
work is equal to the mass times the acceleration
times the displacement. I want to write this equation
in terms of the velocities and not the acceleration
times the displacement. So I'm going to ask you
recall a 1-D kinematics equation that looked like this. The final velocity squared
equals the initial velocity squared plus 2 times
the acceleration times the displacement. In order to use this
kinematic formula, we've got to assume that the
acceleration is constant, which means we're assuming that
the net force on this object is constant. Even though it seems like we're
making a lot of assumptions here, getting rid
of the cosine theta and assuming the
forces are constant, none of those
assumptions are actually required to derive the
result we're going to attain. They just make this
derivation a lot simpler. So looking at this
kinematic formula, we see that it also
has acceleration times displacement. So I'm just going to
isolate the acceleration times the displacement on
one side of the equation and I get that a times
d equals v final squared minus v initial
squared divided by 2. Since this is what
a times d equals, I can replace the a times
d in my net work formula. And I find that the net
work is equal to the mass times the quantity v
final squared minus v initial squared divided by 2. If I multiply the terms
in this expression, I get that the net work
is equal to 1/2 mass times the final velocity
squared minus 1/2 mass times the initial
velocity squared. In other words, the
net work or total work is equal to the
difference between the final and initial
values of 1/2 mv squared. This quantity 1/2
m times v squared is what we call the kinetic
energy of the object. So you'll often hear that the
net work done on an object is equal to the change in the
kinetic energy of that object. And this expression is
often called the work energy principle, since it relates
the net work done on an object to the kinetic energy gained
or lost by that object. If the net work
done is positive, the kinetic energy
is going to increase and the object's
going to speed up. If the net work done on
an object is negative, the kinetic energy
of that object is going to decrease, which
means it's going to slow down. And if the net work
done on an object is 0, it means the kinetic
energy of that object is going to stay the same,
which means the object maintains a constant speed.