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### 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

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# Work and the work-energy principle

Physicists define work as the amount of energy transferred by a force. Learn about the formula for calculating work, and how this relates to the work-energy principle, which states that the net work done on an object is equal to the change in its kinetic energy. Created by David SantoPietro.

## Want to join the conversation?

- Why does one need a cos and theta when we can easily get the work by multiplying F and displacement ?

- thoroughly confused(13 votes)- Imagine that your F is at and angle starting at the origin and pointing somewhere between the x and y axis and you distance is along the x axis. To calculate the word done we need to take the x component of the our F and multiply it by the distance. And easy way to do this is multiplying the cosine of the angle and the F. And only the x component of our force matter because the y component is going vertical and doesn't do anything to with the x-axis movement. If that doesn't make sense imagine F is the hypotenuse of your triangle. And the x-component is the base and the y-component is the height. Knowing the angle between the x-axis and our hypotenuse (F) and using our handy knowledge of trigonometry, we know we can use the cosine of our angle multiplied by our hypotenuse (F) to find the x-component because of the intrinsic ration in triangles. I hope this helps.(29 votes)

- Hi, i'm unclear about why Wnet=0 indicates that an object is maintained at constant. Could someone please kindly help to explain the relationship for this?

The reason for having this doubt:

If Wnet=0, doesn't it mean that the object is not moving at all? Since Wnet also = Fnet(d), it could mean that d=0. Hence there is no work done.

If looking at the application of the work energy theorem whereby acceleration is assumed to be constant & Wnet=1/2(m)(vnet^2) with Wnet=0, it would mean that the net velocity = 0 as the mass of an object will not change. Hence, isn't it that the object will not be moving, instead of moving at a constant velocity?

Please kindly help out in clarifying my understanding. Thank you.(9 votes)- .. the condition is that if the object is moving at a constant speed/velocity then the net work will be zero is TRUE. this is bcoz when u move at a constant velocity then the acceleration is zero bcoz u maintain ur speed instead of increasing or decreasing ur speed . it leads to zero acceleration and NOT constant acceleration. force is mass times acceleration. since u have a zero acceleration your force is going to be zero . and since work is force times displacement ur work will also be zero.

the possibility raised by u that d = 0 is true and work will also be 0.

if the acceleration is constant it means that ur velocity keeps increasing by equal amounts in equal intervals of time.... hence it cannot be zero. i hope that clarifies......if u still have any doubt plz ask.(26 votes)

- Hi,

Just trying to understand Kinetic energy. If a mass travels at a constant velocity in one direction and then moves in the opposite direction at the same velocity will the kinetic energy remain the same?(8 votes)- Wow thats a great question....

How to figure out the answer?

First thing that comes to mind is that energy is not a vector. This means that, whatever the direction, it will have the same value.

Even though velocity is part of the equation for KE; because it is squared... it loses its vector qualities.(23 votes)

- I don't understand 2 things: 1) What are the conditions for when this formula is valid? 2) How does potential energy relate? What if an object is lifted straight up onto a table and placed there? The energy of the system has changed, even if the starting and ending kinetic energies are the same.

I'd learned that change in energy = deltaKE + deltaPE + Wdoneonsystem, or if all work is done on the system (none by the system), change in energy = Wdone= deltaKE + deltaPE.(7 votes)- Positive work is done on the object by the lifter, but negative work is done on the object by gravity. Therefore, the net work done on the object is zero.(4 votes)

- Gravitational Potential Energy does negative work right?

So why isn't it taking away energy from an object which goes up to a certain height? As told in the previous lecture, the object possesses Potential energy. I dont understand ... why does it have energy at all?!(4 votes)- To calculate Grav. potential energy we use Ep = -G (m1m2/r).

In order to have 0 grav. potential energy, the denominator, r, would need to be really massive, so that you were "out of the grav, field." But Newton says that grav, fields extend throughout the universe (even though I am hundreds thousands if not millions of miles away from Pluto its mass and my mass are interacting via our respective grav fields.) So it stands to reason that if you are moving something away from an object, you are moving it towards infinity, so towards 0 grav. potential. But if you let go, it converts the energy you supplied it to kinetic energy, so the conservation of energy says that it must have less energy then it did before you let go. In order for it to have less energy than 0, it must be negative.

It has energy at all because it is still in the gravitational field of the larger object.

Hope this helps!(8 votes)

- What is the exact meaning of 'NET' Force and 'NET' Work?(6 votes)
- 'Net' is used as opposed to 'Gross' because in some systems you get work back. Depending on your sign convention, negative work might be work returning to the system, and once you take all of the incoming and outgoing quantities into account, you have the net work.(4 votes)

- Where does 1/2 come from?(2 votes)
- If you mean the 1/2 from mv²/2 , its a factor that appears when you calculate W = ∫F.ds = m.∫a.ds = m∫(dv/dt).ds = m∫dv.(ds/dt) = m∫v.dv = mv²/2(6 votes)

- why is the cosine theta needed in some cases, and not in other cases? I noticed that cosine theta was taken out during the net work formula, and wanted to know why? Thank you!(3 votes)
- Cosine theta is applied in every case of "work" because it determines whether the system energy is positive, negative, or 0 (neutral). After that the term is used, there is no reason to have it in anymore, because it fulfilled it's one purpose, to tell you the amount of energy put into the system. In conclusion, cosine theta is essential to every "work" problem, but once you determine the direction of the force and displacement, it is just taken out of the equation once you use it because it is not needed anymore for any purpose. The reason why David took it out of the net work formula was because he determined the directions of the displacement and force.(2 votes)

- A man performs 120J of work in 6 seconds .calculate power?(1 vote)
- Why do we need theta when we are finding work as by multiplying force and displacement we can easily find it?(2 votes)
- You can't simply multiply force and displacement unless the force is parallel to the displacement. You need the component of the force that is parallel to the displacement. That's what cos(theta) does.(4 votes)

## 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.