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### Course: AP®︎/College Physics 1 > Unit 5

Lesson 7: Power# Power

In physics, power is defined as the rate at which work is done. In other words, it measures how quickly energy is being transferred or transformed. Explore the concept of power in physics through an example of two weightlifters, one who lifts faster than the other, to see that power measures the rate at which work is done. Finally, learn how to calculate both average and instantaneous power. Created by David SantoPietro.

## Want to join the conversation?

- If power is Force times velocity, and Force is mass times acceleration, then what in the world do we mean by acceleration times velocity. because if there is acceleration, then there is no constant velocity to multiply by... pls help, and thanks(22 votes)
- Well I might be wrong, but I don't understand very well that relation you have done. Firstly,
*instantaneous*power is equal to F · v. If we said that F = m · a, and plug that into the*instantaneous*power, then we would have P = m · a · v. So now we can see that for the power, the mass does matter. Then, the instantaneous speed is a magnitude that is constant by the simple meaning that is in an infinitesimal time interval. So what does this result mean? Well, you are saying that the mass of that object times the total*constant*acceleration it has times the*instantaneous*velocity will result in the power. That particular velocity it has in that particular frame of time does matter to the power the object has. If we were talking about a non-constant acceleration object, then the same analogy could be done. It is important to determine the point at that infinitesimal time interval to decide how much power the object has. Thinking of a car makes the process somehow easier.

As a conclusion, please take this with a grain of salt. I'm not certainly sure if what I'm saying is right. Feel free to correct me or to give a better explanation. Cheers.(5 votes)

- so those people are easily lifting 220 lbs. My boys are strong(14 votes)
- so do u have to always include the cosO in the formula?(5 votes)
- Yes, unless the force is along the axis of displacement. (In other words, say you're pushing an object rightward on a flat surface and it moves to the right. That is moving along the axis of displacement. But say that you're pushing the object to the right up a ramp: the force applied is to the right, but the displacement is along the ramp, which is both right AND up. You therefore introduce the cos-theta.)

Technically, the cosine function is always there, but if difference between the axis of force applied (where you push) and axis of displacement (where what you push goes) is zero, cos(0) = 1, so W = Fdcos(0) is just W = Fd*1 or W = Fd. It just simplifies nicely like that, so you never need to technically calculate the cosine function with such a situation.(14 votes)

- At0:50I'm sort of confused to the fact that the force applied by both weight lifters is the same. For example, I have two robots lifting boxes. Robot 1 moves a 100 kg box 5 meters in 1 second. Robot 2 moves 100 kg box 5 meters in 2 seconds. In order for something to accelerate faster, would more force not have to be applied to the box?

Thanks in advance guys!(6 votes)- Actually, when we say that the force applied on the box by the weightlifters is mg = 980N, that is actually the average force applied by them. The one who lifted it faster might have applied a lot of force at the beginning to get the weight moving faster upwards and then might have just let the weight ascend (or just applied a little force) but the other person isn't strong enough to do that. He could overcome gravity to lift the weight up. We know that the average force exerted by them is 980N because the object starts and ends with no velocity, so the average force has to be equal to gravity. If it were more, the weight would end up with some velocity upwards.(1 vote)

- At0:40, when it's talking about the work done by the weightlifter, what if velocity increases as the weight is lifted. How would you calculate the power then? Would the work be the same? Thanks!(3 votes)
- Hi, power can also be defined as P=Fv, the work would be the same as long as the forces acting on the object are conservative, e.g. gravitational force. If there are non-conservative forces also acting on the object, e.g. frictional force then the work done would not be the same. I hope that helps to answer your question.(6 votes)

- So the Work is synonymous with potential energy and kinetic energy ?(2 votes)
- well, work need not always be mechanical energy[PE+KE], it could be gravitational energy, chemical energy, electrical energy etc.(1 vote)

- What about kinetic energy? If the guy on the right is moving his weight at a faster velocity, wouldn't he have a greater KE? I would think that he'd also would have done more work, though I do understand that its the same because of the PE equation. But can someone explain this in terms of the KE equation?(3 votes)
- If the object is lifted faster, the lifter either has to exert more negative work to stop the object at the top, or exert less work after the initial motion because the gravitational force will slow down the upwards motion for the lifter. In both cases, the same work will have been exerted on the object: net work = ΔKE = 0 (the object is still at the beginning and continues to be still at the end.)(2 votes)

- can we use p=fv for average power?(2 votes)
- depends on the situation. But, in general, it is not an equation for average power.

if, for example, during a certain time period, the velocity or the force change value, then the power changes value too.(3 votes)

- What if there are muliple forces acting on the body?

What will you substitute for F?(2 votes)- With multiple forces you usually use Net Force which is basically the vector sum of all of the forces.(3 votes)

- I was thinking of this in the context of derivatives and arrived at another question:

Since`W=∫ᵢᵉ F dt = Eₑ - Eᵢ`

, and`P = dW/dt = d/dt[∫ᵢᵉ F dt] = d/dt[Eₑ - Eᵢ] = Eₑ*dEₑ/dt - Eᵢ*dEᵢ/dt`

I was thinking, if the derivative of velocity, whose units is m/s, is m/s², then what is the "derivative unit" of N*m=J?(2 votes)- To quickly answer your question, the derivative unit of N*m (Joule) would be a N*m/s, which can be simplified as a Joule/s or, simply, a Watt.

Above, you're defining work incorrectly by your integral. Under the assumption that the force is constant, and it is parallel (0 degrees, or in the same direction) to the displacement, work should be defined as W = Force*displacement, or W=F*s, where s is displacement.

You showed above that the units for Work are units of energy, which are Joules. If you take the derivative of these units, d/dt(Work), you will get J/s or Watts. Just think of the denominator in the d/dt as adding in an infinitesimal amount of time in seconds to the denominator of the variable you are taking the derivative of.(2 votes)

## Video transcript

Check out these weightlifters. The one on the right is
lifting his weight faster, but they're both doing
the same amount of work. The reason I can say
that is because work is the amount of energy
that's transferred. Or to put it a
simpler way, this is the way I like to
think about it, work is equal to
the amount of energy you give something or
take away from something. Both weightlifters are
giving their weights the same amount of
gravitational potential energy. They both lift them two
meters, and the masses are 100 kilograms each. Plug those into the formula for
gravitational potential energy, and you find that the work
done by each weightlifter is 1,960 joules. But the weightlifter
on the right is lifting his weight faster. And there should be
a way to distinguish between what he's doing
and what the other slower weightlifter is doing. We can distinguish
their actions in physics by talking about power. Power measures the
rate at which someone like these weightlifters or
something like an automobile engine does work. To be specific, power is
defined as the work done divided by the time that
it took to do that work. We already said that
both weightlifters are doing 1,960 joules of work. The weightlifter on the
right takes 1 second to lift his weights, and
the weightlifter on the left takes 3 seconds to
lift his weights. If we plug those times into
the definition of power, we'll find that the power
output of the weightlifter on the right during his lift
is 1,960 joules per second. And the power output
of the weightlifter on the left during his lift
is 653 joules per second. A joule per second
is named a watt, after the Scottish
engineer James Watt. And the watt is abbreviated
with a capital W. All right, let's look
at another example. Let's say a 1,000 kilogram
car starts from rest and takes 2 seconds to reach a
speed of 5 meters per second. We can find the power
output by the engine by taking the work
done on the car divided by the time it took
to do that work. To find the work
done on the car, we just need to figure
out how much energy was given to the car. In this case, the car
was given kinetic energy and it took two seconds to
give it that kinetic energy. If we plug in the values
for the mass and the speed, we find the engine had a
power output of 6,250 watts. We should be clear that what
we've really been finding here is the average power
output because we've been looking at the total
work done over a given time interval. If we were to look at
the time intervals that got smaller and smaller, we'd
be getting closer and closer to the power output
at a given moment. And if we were to make our time
interval infinitesimally small, we'd be finding the power
output at that particular point in time. We call this the
instantaneous power. Dealing with
infinitesimals typically requires the use of
calculus, but there are ways of finding
the instantaneous power without having to use calculus. For instance, let's
say you were looking at a car whose
instantaneous power output was 6,250 watts at
every given moment. Since the instantaneous
power never changes, the average power just equals
the instantaneous power, which equals 6,250 watts. In other words, the average
power over any time interval is going to equal the
instantaneous power at any moment. And that means work
per time gives you both the average power and
the instantaneous power in this case. Let's say you weren't so lucky,
and the instantaneous power was changing as
the car progressed. Then, how would you find
the instantaneous power? Well, we know that power
is just the work per time. So something we
can try is to plug in the formula for work, which
looks like FD cosine theta, and then divide by the time. Something that you might
notice is that now we have distance per
time in this formula. So let's isolate the
distance per time. Distance per time
is just the speed. So I can replace d over
t with v in this formula. And if you plug in the
instantaneous speed of the car at a given moment
in time, you'll be finding the
instantaneous power output by the force on the car at
that particular moment in time. So to find the instantaneous
power output by a force, plug in the force on the object
at a particular moment in time, multiply by the
speed of the object at that same moment in time,
then multiply by cosine theta. But be careful here. Theta isn't any old angle. It's the angle between
the force on the object and the velocity of the object. But in many cases, the force
is in the same direction as the velocity, which means
the angle between the force and the velocity is zero. And since cosine of 0
is 1, you don't really need the cosine in
the formula at all. And you find that the
instantaneous power is just the force times the speed. All right. So what does power mean? Power is the rate at
which work is done. What does average power mean? Average power is the work done
divided by the time interval that it took to do that work. What does the
instantaneous power mean? Instantaneous power
is the power output of a force at a
particular moment in time.