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# 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.

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• 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
• 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.
• so those people are easily lifting 220 lbs. My boys are strong
• so do u have to always include the cosO in the formula?
• 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.
• At I'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?

• 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)
• At , 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!
• 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.
• So the Work is synonymous with potential energy and kinetic energy ?
• 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?
• 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.)
• can we use p=fv for average power?
• 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.
• What if there are muliple forces acting on the body?
What will you substitute for F?
• With multiple forces you usually use Net Force which is basically the vector sum of all of the forces.
• 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?