Main content

### Course: MCAT > Unit 9

Lesson 5: Work and energy- Work and energy questions
- Introduction to work and energy
- Work and energy (part 2)
- Work and the work-energy principle
- Work as the transfer of energy
- Work example problems
- Work can be negative!
- Conservation of energy
- Work/energy problem with friction
- Intro to springs and Hooke's law
- Potential energy stored in a spring
- Spring potential energy example (mistake in math)
- Conservative forces
- Power
- Introduction to mechanical advantage

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Work and energy (part 2)

Continue to explore the relationship between work and kinetic energy, seeing how applying force to an object over a distance generates kinetic energy. You'll also be introduced to the concept of potential energy, seeing how work done to lift an object creates gravitational potential energy. Finally, you'll get a hint towards the law of conservation of energy. Created by Sal Khan.

## Want to join the conversation?

- I didn't understand what Sal said at3:35. If he is applying the same amount of force as force due to gravity and both forces are equal and opposite in direction

why is the object moving at all? Can someone explain?(21 votes)- Hello Maha, Brilliant Quesiton, i was once in your shoes.

Here is my simple explanation. remember that F=ma. a stands for acceleration. not velocity. So as low as the object is moving with constant velocity. it means that acceleration equals to 0 (because there is no change in velocity) and so the net force is also equal to zero.

take these two examples. You are sitting in a chair, your weight force is equal to reaction force and thus you don't fall through the ground. happy? good.

second example is you go sky diving (great experience, i would highly recommend it), and you accelerate due to gravity. but after some time you reach terminal velocity. that is when the weight force equals to the air friction force. so net force = 0. But you don't stop moving. you just move with constant velocity.

These ideas link with newton's laws. Inertia is a important concept you need to get down.

whenever you get confused with physics in general, i would look at the equations and try to understand and visualize what are they saying. it helped me quite a bit

i hope this helps, best of luck(20 votes)

- what is actually the diference between speed and velocity?(1 vote)
- Speed needs only magnitude to define.Velocity needs magnitude as well as direction to define.Speed is a scalar quantity. Velocity is a vector quantity.(17 votes)

- If potential energy is the ability to do work (7:00), what work does the object do when it falls? Isn't this work (falling of the object) done by gravity?(7 votes)
- Remember, gravity is an acceleration (9.81m/s^2); it's not a force.

Therefore, 10kg of an object has potential to do work WITH GRAVITY, if it's located at proper position.(4 votes)

- I know g= 9.8 m/s or rounded to 10 m/s. When do you use either number? Because in this video, in the first example 9.8 is used but in the second example 10 is used. I'm not sure what constituted the use of either numbers in the specific problems.(1 vote)
- 9.8 is more accurate than 10, but the numbers are not what is important. The key to physics is understanding the principles and how to apply them.(7 votes)

- At5:50, if F=ma, and the elevator isn't accelerating, why doesn't F=0 and thus work=0? I get, intuitively, why there is work being done, but based on the equations why does it not work out like that?(4 votes)
- F does = 0

The work done by the elevator is equal to the (negative) work done by gravity. So the net work is 0.

The work done by the elevator is positive and it appears as an increase in the PE(2 votes)

- if the force pulling upwards on the elevator is equal to the force pulling downwards , wouldnt that mean that the elevator wouldnt move? since there are two forces of equal proportion acting on it , from different sides.(2 votes)
- No. Balanced forces mean the elevator won't accelerate. Constant velocity = no acceleration. The constant may be 0, but it doesn't have to be.(6 votes)

**bold**Question. What does all this mean(3 votes)- so,potential energy only to upwards?how about downwards,is there a potential energy?(1 vote)
- Potential energy is ANYTHING that stores energy. This could be in a battery, a rubber band, or gravity. Gravitational potential energy is only moving something up, but other kinds of PE can be in any direction, or even stationary.(3 votes)

- K.E = (mv^2)/2

final velocity = v

initial velocity = u

time = t

mass = m

acceleration = a

But,when I do this:

Work = F * displacement = K.E

⇒ K.E = F * vt ...[∵ displacement = velocity *time]

⇒ K.E = ma * vt ...[∵ F=ma]

⇒ K.E = m((v-u)/t) * vt ...[∵ mass = (v-u)/t]

⇒ K.E = m(v - u) * v ...[the t's cancel out]

⇒ K.E = m(v) * v ...[∵ the body is at rest, initial

velocity is zero and therefore

v-u = v]

Therefore, K.E = mv^2

But,Kinetic energy(K.E) is supposed to be equal to:

K.E = (mv^2)/2

What is wrong with my derivation?Why don't I get :

K.E = (mv^2)/2

Please answer, cause if you don't, all the effort I put into typing this would go to a waste.(2 votes)- You interpreted displacement incorrectly. Displacement = average velocity * time

Assuming constant acceleration, average velocity = 1/2(final velocity - initial velocity)(3 votes)

- A place where there is no gravity will a body posses potential energy there? if not what will happen if a rubber band is stretched there and given a certain configuration, won 't it posses potential energy?(1 vote)
- You are confusing gravitational potential energy with elastic potential energy. The stretched rubber band has elastic PE as a result of being stretched. This is true regardless of whether it is in a gravitational field or not.(5 votes)

## Video transcript

Welcome back. In the last video, I showed you
or hopefully, I did show you that if I apply a force
of F to a stationary, an initially stationary object with
mass m, and I apply that force for distance d, that that
force times distance, the force times the distance that
I'm pushing the object is equal to 1/2 mv squared, where
m is the mass of the object, and v is the velocity of the
object after pushing it for a distance of d. And we defined in that
last video, we just said this is work. Force times distance by
definition, is work. And 1/2 mv squared, I said this
is called kinetic energy. And so, by definition, kinetic
energy is the amount of work-- and I mean this is the
definition right here. It's the amount of work you need
to put into an object or apply to an object to
get it from rest to its current velocity. So its velocity over here. So let's just say I looked at an
object here with mass m and it was moving with
the velocity v. I would say well, this
has a kinetic energy of 1/2 mv squared. And if the numbers are confusing
you, let's say the mass was-- I don't know. Let's say this was a 5 kilogram
object and it's moving at 7 meters per second. So I would say the kinetic
energy of this object is going to be 5-- 1/2 times the mass
times 5 times 7 squared, times velocity squared. It's times 49. So let's see. 1/2 times 49, that's
a little under 25. So it'll be approximately 125
Newton meters, which is approximately-- and Newton
meter is just a joule-- 125 joules. So this is if we actually
put numbers in. And so when we immediately know
this, even if we didn't know what happened, how did this
object get to this speed? Let's say we didn't know that
someone else had applied a force of m for a distance of
d to this object, just by calculating its kinetic energy
as 125 joules, we immediately know that that's the amount of
work that was necessary. And we don't know if this is
exactly how this object got to this velocity, but we know that
that is the amount of work that was necessary to
accelerate the object to this velocity of 7 meters
per second. So let's give another example. And instead of this time just
pushing something in a horizontal direction and
accelerating it, I'm going to show you an example we're going
to push something up, but its velocity really
isn't going to change. Invert. Let's say I have a different
situation, and we're on this planet, we're not
in deep space. And I have a mass of m and
I were to apply a force. So let's say the force that I
apply is equal to mass times the acceleration of gravity. Mass times-- let's just call
that gravity, right? 9.8 meters per second squared. And I were to apply this force
for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times
the distance is equal to-- well the force is mass
times the acceleration of gravity, right? And remember, I'm pushing with
the acceleration of gravity upwards, while the acceleration
of gravity is pulling downwards. So the force is mass times
gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm
applying is going to be h. And what's interesting is-- I
mean if you want to think of an exact situation, imagine an
elevator that is already moving because you would
actually have to apply a force slightly larger than the
acceleration of gravity just to get the object moving. But let's say that
the object is already at constant velocity. Let's say it's an elevator. And it is just going up with
a constant velocity. And let's say the mass of the
elevator is-- I don't know-- 10 kilograms. And it moves up
with a constant velocity. It moves up 100 meters. So we know that the work done
by whatever was pulling on this elevator, it probably was
the tension in this wire that was pulling up on the elevator,
but we know that the work done is the force necessary
to pull up on it. Well that's just going to
be the force of gravity. So we're assuming that
the elevator's not accelerating, right? Because if the elevator was
accelerating upwards, then the force applied to it
would be more than the force of gravity. And if the elevator was
accelerating downwards, or if it was slowing down upwards,
then the force being applied would be less than the
acceleration of gravity. But since the elevator is at a
constant velocity moving up, we know that the force pulling
upwards is completely equal to the force pulling downwards,
right? No net force. Because gravity and this force
are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward
force is equal to the force of gravity. At least in magnitude in
the opposite direction. So this is mg. So what's m? m is 10 kilograms.
Times the acceleration of gravity. Let's say that's 9.8 meters
per second squared. I'm not writing the units here,
but we're all assuming kilograms and meters
per second squared. And we're moving that for a
distance of 100 meters. So how much work was put into
this elevator, or into this object-- it doesn't have to be
an elevator-- by whatever force that was essentially
pushing up on it or pulling up on it? And so, let's see. This would be 98 times 100. So it's 9,800 Newton meters
or 9,800 joules. After we've moved up 100 meters,
notice there's no change in velocity. So the question is, where
did all that work get put into the object? And the answer here is, is that
the work got transferred to something called
potential energy. And potential energy is
defined as-- well, gravitational potential
energy. We'll work with other types of
potential energy later with springs and things. Potential energy is defined
as mass times the force of gravity times the height
that the object is at. And why is this called
potential energy? Because at this point, the
energy-- work had to be put into the object to get it
to this-- in the case of gravitational potential energy,
work had to be put into the object to get
it to this height. But the object now, it's not
moving or anything, so it doesn't have any
kinetic energy. But it now has a lot of
potential to do work. And what do I mean by potential
to do work? Well after I move an object up
100 meters into the air, what's its potential
to do work? Well, I could just let go of it
and have no outside force other than gravity. The gravity will
still be there. And because of gravity, the
object will come down and be at a very, very fast velocity
when it lands. And maybe I could apply this to
some machine or something, and this thing could do work. And if that's a little
confusing, let me give you an example. It all works together
with our-- So let's say I have an object
that is-- oh, I don't know-- a 1 kilogram object and
we're on earth. And let's say that is 10 meters
above the ground. So we know that its potential
energy is equal to mass times gravitational acceleration
times height. So mass is 1. Let's just say gravitational
acceleration is 10 meters per second squared. Times 10 meters per
second squared. Times 10 meters, which
is the height. So it's approximately equal to
100 Newton meters, which is the same thing is 100 joules. Fair enough. And what do we know
about this? We know that it would take about
100-- or exactly-- 100 joules of work to get this
object from the ground to this point up here. Now what we can do now is use
our traditional kinematics formulas to figure out, well, if
I just let this object go, how fast will it be when
it hits the ground? And we could do that, but
what I'll show you is even a faster way. And this is where all of
the work and energy really becomes useful. We have something called the law
of conservation of energy. It's that energy cannot be
created or destroyed, it just gets transferred from
one form to another. And there's some minor
caveats to that. But for our purposes, we'll
just stick with that. So in the situation where I just
take the object and I let go up here, up here it has a
ton of potential energy. And by the time it's down here,
it has no potential energy because the height
becomes 0, right? So here, potential energy is
equal to 100 and here, potential energy
is equal to 0. And so the natural question is--
I just told you the law of conservation of energy, but
if you look at this example, all the potential energy
just disappeared. And it looks like I'm running
out of time, but what I'll show you in the next video is
that that potential energy gets converted into another
type of energy. And I think you might be able
to guess what type that is because this object is going to
be moving really fast right before it hits the ground. I'll see you in the
next video.