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Why is the concept of Power useful?

  1. Apr 1, 2010 #1
    Can someone explain to me please why this concept is useful ?

    I can't make sense out of it, am struggling...

    Thank you
  2. jcsd
  3. Apr 1, 2010 #2


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    It tells you how big of an engine a car needs and how useful an exercise bike is for losing weight.
  4. Apr 1, 2010 #3
    Also, it measures how much you spend each day on your electricity! :)
  5. Apr 2, 2010 #4


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    Because power is the rate at which you use energy, and energy is the ability to do work.
  6. Apr 2, 2010 #5
    Power is simply the rate at which work is being done. It is a useful definition in physics since it has a time component to it.
  7. Apr 2, 2010 #6
    If you want to get something done in a certain time, then power is useful...because that's pretty much what power is (how much and how fast you can get done).

    If you don't care how long it takes to get something done, then it's not so useful. For example, if you want to walk to your friend's house, but don't care if it takes you 1 hour or a lifetime, then power is indeed useless.

    But realistically, most things that people want to get done, they want to get them done within their lifetime. That's when the concept of power is useful. I'm talking anything that requires work, anything that can be described with a verb: building, lifting, moving, breaking, spinning, opening, pumping, accelerating, slowing, tilting, turning, closing, heating, smashing, squeezing, throwing, etc...
    Last edited: Apr 2, 2010
  8. Apr 2, 2010 #7
    In physics it's useful because it relates two of the most important concepts; energy and time.
    Unfortunately the word "power" also has an everyday usage which is different from the scientific usage. Here I mean its use in expressions such as political power, willpower etc.
    Here the concept is more vague, and consequently less useful to a physicist!
  9. Apr 2, 2010 #8
    One thing i dont understand is........

    if a vehicle is said to have a power of 15kW, does that mean the power is the power considering resistance forces or without ?
    or is this 15kw the work it does in a certain condition only? and it can do more work under different conditions? afterall power is just the rate it does work and not a unchangeable value, right?

    is power defined depensing on say the resistance? because if a machine pushes a stone for 10m in 5s at 20N and resistance is 10N, net force is 10N which gives net work = 100J and power = 20W.

    so these 20W is the power considering resistance.

    now the actual work done by the force is 20N * 10m = 200J right? and power = 40W.

    which one is correct ??

    afterall a machine can work under different circumstances

    if it does then 15kJ a second, this force moving it is the force it generates minus resistance or is it the total force it does disregarding resistance?

    thank you
    Last edited: Apr 2, 2010
  10. Apr 2, 2010 #9
    These are important questions to ask, and asking leads to understanding.

    Power is defined in two useful ways, the second of which can be rigorously derived from the fundamental theorem of calculus:

    1. Power = Energy (or Work, if you prefer) / time

    2. Power = Force * Velocity. More specifically, that is:
    The power of a machine, say, a rock-mover =
    The product of the Force it can apply to a rock times the velocity that rock might be moving at.

    Does power change with resistance? No. Not resistance. But it DOES change with the velocity of the object being moved. This is best explained with an example:

    Suppose you decide you're gonna start riding your bike. So you get on, you push the pedals with all your might, but damnit, you can't get the thing to move because the gear is too high. Well, elementary physics teacher says, you ain't doin' no work. No work means no power.

    However, once you get the bike moving along, you are perfectly capable of pumping and moving yourself along. Now look: You're clearly applying a force to the pedals, and the pedals are moving. By my "Definition 2" of power, your legs are now powerful!

    Power IS a useless concept WHEN THE BIKE IS AT REST. Clearly, the following equation is not valid:


    However, once the bike is moving, power is VERY useful, ESPECIALLY for calculating your maximum velocity. Suppose the force of all forms of friction is known to be "f" Suppose, then, that you pedal, and go as fast as you possibly can. You will eventually find that you are reaching a maximum speed. Once you reach this speed, then we can say that your velocity is constant. A constant velocity means sum of the forces is zero, and that means that the motive force of your legs equals the opposing force of friction, "f." Now it IS valid to say that the power of your legs P = f * v where v is the maximum velocity you were able to achieve riding on a surface with friction force f.

    Now the concept of power is useful, when, say, you move onto the grass where the force of friction is "f-grass." Suppose you want to know how fast you are capable of riding on the grass. I will be able to INSTANTLY tell you that

    Maximum velocity = P / "f-grass" ​

    where P is the SAME constant we calculated on the pavement.

    If you're willing to read on, I'll tell you now what's going on in your situation. If the machine "pushes a stone for 10m in 5s at 20N" then the MACHINE is performing 200 J of work regardless of the resistance. The power of the machine is NOT concerned with the resisting force. However, since the rock is accelerating, that means its velocity is increasing, so, technically speaking, the power of the machine is increasing the whole time. When you think about it, it also doesn't make sense that the machine should always be able to apply this force of 20 N to the rock, as then it could accelerate that rock to the speed of light! The point is, it's a poorly stated problem.

    My final answer: the power of the machine is NOT constant over the interval t=0 to t=5. The power would increase over the interval. Since velocity is not given, the power cannot be calculated.

    Keep in mind that you did make a mistake. You claimed that the power of the machine can be determined by the NET force on the rock. That is not true. The power of the machine is calculated from the force which the machine itself applies to the rock.

    I hope this helps.

  11. Apr 3, 2010 #10


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    You'll have to ask the person who said it what they mean. Context matters.
  12. Apr 3, 2010 #11


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    Unless of course you are talking about electrical power:

    P=I2R = V2/R
  13. Apr 3, 2010 #12

    Jake, your explanation was beautiful, I thank you very much. I was getting to that point here and your explanation made it much faster.

    I would just like to quote this:

    I haven't yet done any calculation to understand this, but my intuition gets me confused.

    By all friction forces are you including mostly air resistance?

    Let's say I have a force of 50N pushing a body on a smooth horizontal surface, but air resistance is there, and it's increasing as speed increases, right? so at a point air resistance will balance my force and hence i will stay at a constant speed V, so the power P = 50*V joules.

    Now talking about the grass, lets say it has a resistance of 10N, so my net force is 40N, now... whether I am using 40N or 50N, my final speed is going to be the same, as the total air resistance at the end will be the same, only the time to get to the final speed will increase, is that correct?

    and so my maximum speed on grass should be the same as on the smooth surface, no?

    and also my force is the same, so whats my power on the grass? P = (Friction + Air Resistance) * Final Speed, where this final speed is the same as before?

    and friction + air resistance should also equal my 50N

    so I am getting it wrong then, i am totally confused now.

    Could you help me please?

    Thank you very much
    Last edited: Apr 3, 2010
  14. Apr 3, 2010 #13
    Okay, I will create a problem, and present its solution:

    Suppose Constantino is capable of riding his bike at 15 meters/second against a constant friction force (no air resistance) of 20 Newtons of force. How fast would Constantino be able to ride his bike if an air resistance constant of k = 0.2 were introduced? In this problem, air resistance force is proportional to velocity (as opposed to the square, which is more realistic)


    The first situation is a test-situation in which we can find out how powerful Constantino's legs are. If his max speed is 15 m/s, then we use P = F * v to find that Constantino's legs can generate 300 W of power when riding his bike at full speed, under regular conditions.

    [tex]P = 300W[/tex]​

    Now we must set up the equation, which we plan to solve for vmax:

    [tex]P = F \cdot v[/tex]​

    The power of his legs will be the same. I repeat, changing what kinds of resistance are there will NOT change the power of his legs. Therefore, P = 300W. Also, substituting the fact that F = kv + ffriction (and the force of regular friction is still 20 N, and air resistance is proportional to v) we have:

    [tex]P = (kv + f)*v[/tex]
    where f means "force of friction" and v means "maximum velocity" (i.e, the solution to the problem!)​

    Then, the quadratic equation reveals itself, after routine algebra.

    [tex]kv^2 + fv - P = 0[/tex]

    [tex]v = \frac{-f \pm \sqrt{f^2 + 4kP}}{2k}[/tex]​

    And plugging in, the fun part...

    [tex] v = 3.62 m/s [/tex]​

    ==================== QED (?) ===========================

    Note that the units of k here are inverse seconds. You can verify that the quadratic equation, as well as the application of the quadratic formula, all have proper units.

    Be careful that you phrase problems VERY clearly. If you feel a problem has not been stated clearly, restate it to yourself.

    As a segway toward responding to OmCheeto's comment about electrical resistance, in which he points out that P = I^2*R, let me point out:

    I learned about power during an experiment for my high school mechanics class. My teacher asked me to find the correlation between the voltage on an electric motor, and the power generated by the electric motor. I realize this may not be the best experiment, but I attached a metal cylinder (whose moment of inertia I could calculate easily and accurately), then simply measured how quickly the rotational kinetic energy of the cylinder increased by measuring its rotations/second (Which is equivalent to measuring the rate at which the circuit is delivering energy to the cylinder). While the proportion

    [tex]Power \propto Voltage[/tex]​

    was true (and mathematically, the speed of the cylinder increased approximately proportional to the square root of time--do the calculations, you'll see why that must be the case for the proportion to be true [note, kinetic energy = 1/2 * I * omega^2, where omega is rotational speed]).

    However, when the cylinder first started spinning, its velocity was increasing linearly!! But still, THINK ABOUT IT! The square root graph, while it makes sense for large values t, does NOT make sense for t near 0! That would imply that the cylinder jerked instantly, and that pretty much defies common sense and experience.

    If you are confused, it may help to note that the torque created by the motor was by no means constant. If it were, then none of those equations would be true.

    But, for your purposes Constantino--the purposes of classical mechanics--bare in mind: Power ALWAYS equals F * v. Therefore, if you want to know the power of something, then the product of Force and velocity should be constant-- otherwise, power isn't constant, and then there's really no point in calculating it.

    Well, I had fun writing this dissertation, and it was a good review for myself. Thanks for reading and thoughtful responses will be diligently read!

    Last edited: Apr 4, 2010
  15. Apr 4, 2010 #14
    Firstly thank you all very much for the explanations.


    Can I please ask you some questions?

    Do other kinds of friction apart from air resistance increase with speed as well or not?

    Or is speed limit just a cause of air resistance?

    Because if so I think it changes all the situation, at least in my mind...

    what I dont get is... if power is force * velocity, why does it make sense to keep/use the same power value? do the speeds on different surfaces change because the net force is different?

    i dont get it but i'd like so much to. been thinking of it everyday all day long...

    if the force i can apply is constant, we consider then the frictional force and subtract it from my force, so my net force which is gonna give me the velocity will decrease.

    the time it takes for air resistance to balance a force F, it will be different because the terminal speed is reached after different amounts of time for different forces right? but how do we calculate this terminal speed ? i know we will accelerate until air resistabce lets us, but until when will air resistance be lower?

    if my speed limit is 50m/s when i have a force of 50N my power is 2500W.

    so if I apply the same force against a different force of friction, how will my speed change??

    since I am against a higher resistance,my net force will be different and hence i will accelerate at a different rate, but what is the time it takes to slow me down, and hence STATE what my final speed is on that surface? because obviously then the time it takes to slow me down will state my final speed as i will accelerate for this time and at a slower and slower rate.

    I think i need to do some calculations and experiences here to understand it.

    so in a few words, my final speed on any surface will be given by (F - R) / M * T, where F = my force, R = friction +air resistance(which increases with time), M = mass, T = time it takes for air resistance to slow me down.

    so a force of 10000N will cause my speed to increase much faster, in much less time than a force of 10N, but as air resistance is proportional to speed, i will only start to slow down when my speed reaches a certain value right? and if so, the time it takes to slow me down is different for each Force, but the speed at which i will start to slow down.. how fast is reached?

    I appreciate your help very much.

    Last edited: Apr 4, 2010
  16. Apr 4, 2010 #15
    Well, none as far as I can think of, but I'm sure there are other situations where resistance is proportional to speed (perhaps raised to some power). I know that mass changes with speed at speeds near the speed of light, that's easy to look up on wikipedia if you're interested, I think.

    You really do ask the right questions. (Like Detective Spooner in I Robot--"That, detective, is the right question...") Okay, so where did I get that assumption? Well, it's pretty much just that--an assumption. You keep asking why? to a physicist (though I am not a real "physicist") and you'll eventually get his tongue tied. But I guess I thought of it this way...every machine--your body, or a rock-mover, whatever-- has some source of energy. For example, your legs use the mitochondria in the cells if you wanna get technical.

    Now obviously, at the sight of the word "mitochondria" any physicist should close his eyes and hope it goes away, because the thought of performing calculations on a system so complex would scare anyone! But my assumption (that your legs should have a limit to how fast they can expend energy) comes from the fact that... Well... I assume your leg muscles can only "burn calories" so fast.

    So, I've said that power is not constant--it changes with velocity. However, I will also say that a machine has a LIMIT to the amount of power it can produce. And that limit, I ASSUME should stay constant. So, I should make a clear distinction. When a machine first starts operating, (say, when you start riding on your bike) it may not be able to operate at "full power." Once it gets operating, it should reach the point (say, when your bike is going at full speed) that you can pretty much safely assume that the machine is operating at "full power." Then, unless the mechanics of that machine change, we can assume that that machine's "full power" should remain constant from one situation to another.

    First of all, stop talking about net force. Let's talk about the force ONLY coming from your machine (your legs).

    Your assumption that your force will be "constant" is wrong! When you begin riding your bike, and you start pushing on the gears with all your might, your force may very well be 1000 or more Newtons on the pedals, but I guarantee you, the force will be less once you're riding. So let's compare our assumptions:

    My assumption: a machine's "full power" will be constant, regardless of situation.

    Your assumption: a machine's force will be constant, regardless of situation.

    Your assumption is wrong. As it turns out, we can rely on this "full power" constant to remain constant BETTER than we can rely on force to be constant. And that, I suppose, is the answer to your question posed in the title of this thread.

    Now do you see why you've been so confused. You're assumption that you SHOULD EVEN HAVE a "constant" force of 50N is WRONG! And it's understandable that you had this assumption since these bogus physics problems somebody gave you are completely misleading.

    Now that you realize that it may very well be possible that any machine (such as your legs on the bike) will NOT have constant force, you should try to reconsider all your questions, and see what you can answer for yourself!

    I should add a disclaimer here: the assumption "Full power remains constant" is an assumption which only remains true when ALL the mechanics of the machine remain the same. Here's an example where this assumption is used too loosely:

    Joe's arm can throw a ball 50 yards. Therefore (after some calculations, blah blah) Joe's arm has power "X." Therefore, if Joe uses a ball thrower (a stick which extends the length of his arm), then since the ball thrower is just a rigid object, it shouldn't contribute any energy to Joe. Therefore, using a ball thrower will NOT increase Joe's power. Therefore, it will not let him throw the ball further....
    WRONG! The ball thrower changed the mechanics of his arm. Just like changing gears on your bike lets you go faster, if you're smart about it. (Here, I'm considering as one mechanical object, You + the Bike). The gears aren't contributing power. That's all (still) just coming from the mitochondria. However, a new mechanical object will have a new "full power."

    Now, by the way, power is not all-powerful. It's a long read, but my thread, "A Paradox Which Must Be Resolved" shows that when you KNOW what forces are involved, using newtons law, f = ma, or f = d/dt [mv] is more reliable, and that Conservation of Energy doesn't work sometimes (when energy may be lost to heat or internal energy). Just skip through the math and read the discussion. It'll show you how you must use great discretion when choosing your method of solving a problem. Make sure you're clear on your assumptions!!

    Last edited: Apr 4, 2010
  17. Apr 6, 2010 #16
    Everyone, thank you very much for your posts.

    After 1 week in a dark cave with only a candle I think I got the concept, I been thinking of it everyday trying to understand the why's. So I was doing some exercises and found one that had a situation that was linked to my thoughts.

    The exercise was simple but I think the concept of was clarifying for me, and it was the following:

    The engine of a van weighting 3 tonnes works at a constant rate. On the level road the maximum uniform speed is 18 km/h but up an incline of 1 in 10 it is 8 km/h. Assuming the resistance to be the same in the two cases, find the power of the engine. Assume g = 10 m/s^2.

    I will solve it here so you can follow my reasoning, I will explain afterwards what the exercise brought to my mind and then you tell me whether it is correct.

    18 km/h = 5m/s
    8 km/h = 20/9 m/s

    So on the level road, the power P = 5F where F is the pull of the engine.

    Assuming g = 10 m/s^2:
    On the incline, resolving the force parallel to the incline, 3000g * 1/10 = 3000N
    Up the incline, the power equation P = (3000 + F) * 20/9


    5F = (3000 + F) * 20/9
    F = 2400N

    So the power P = 12000 watts

    After the solving, the suttle part for me was...

    If the power of the engine is constant, then it does the same amount of work per second, but the speeds are different and the only reason why it is different is because of the gravitational force on the incline.

    On the level, the only resistance force I assume is air resistance, i ignore friction. So the van reaches its speed limit in a certain amount of time.

    On the incline, gravity is pulling the car downwards, but the amount of work per second is the same, but if it is the same, the engine pull should be the same and so should be the speed, but they aren't, and i found that it isn't the same because gravity is doing negative work, and that means there is work leaking off ! So the engine does the same amount of work, but part of this work is being dissipated by gravity.

    And negative work means a negative force through a distance, which is only natural because indeed the net force is less now, which causes speed to be less.

    If I am wrong and am not seeing it, please help!
  18. Apr 6, 2010 #17
    Okay it's a good problem.
    The answer is correct, so let's make sense of the solution, since you're confused.

    This reasoning is correct.

    WHAAAAT?? No! This problem assumes a resistance is "the same" in both situations. It MUST be referring to force of friction F. If the force of resistance were air resistance, then it would have DECREASED in the second situation since the truck is moving slower. However, since we can assume the force is friction, you should know that friction is considered constant. Those people on the other discussion thread (on air resistance) confused you. Don't think about air resistance right now; the problem doesn't talk about it.

    true. Now let's look at that more closely. On the incline. When the car has reached maximum velocity of "V", we know:

    [tex]\sum F_{i}= 0[/tex]

    since the velocity is constant (i.e not accelerating)
    Now, as you correctly pointed out, there are two forces against the car parallel to the incline: mg*sin(theta) and F (the unknown constant resistance force). Since the incline is small, we can approximate sin(theta) = tan(theta) = 1/10, which you already did. Good. But then, we said the sum of the forces is zero, so... the motive force of the engine will balance these forces, and will therefore be 1/10 mg + F. This shouldn't be surprising to you. By now you should know that it's possible that the engine will actually produce different amounts of force. In the first situation it was producing a force exactly F. In the second, it's actually producing a greater force, to counterbalance gravity.

    Well, I don't know what the word "pull" means, but I'm assuming you mean force. And no, it need not be the same. POWER WILL BE THE SAME. Force will vary. Read the above paragraph.

    I don't know what it means if work is leaking off. Please use the terminology accepted by physicists or we won't be able to help you. I believe this paragraph is the result of your previous misconception, namely that the motor will always produce the same force in any situation. But the most egregious error is in the second paragraph there: that net force should be less because the speed is less. Paulo... force = mass * acceleration. You need to understand that equation better before you can move on to discuss power, I think. You also need to understand what acceleration is. It is a change in velocity. Realize that in these situations presented in the problems, the car has already reached its maximum velocity. That velocity is not changing. Therefore it is not accelerating. And therefore, I repeat:

    [tex]\sum F = 0[/tex]

    I think you get the basic concepts, but I think you have trouble applying them to real problems. Keep trying, it'll come with practice.

  19. Apr 7, 2010 #18
    Hey Jake, thank you again for your reply.

    Now this, how can an engine generate more force only because of gravity? it's as if the engine is smart enough to know that gravity is pulling downwards and so it needs to apply more power. Is it really like that?

    I think you are talking about what happens to this equation: P = (3000 + F)20/9

    where 3000 + F is the force, but what I thought of it is that the equation means the amount of work and so, 3000 * 20/9 + (20/9) * F is the total work done per second, and the important part there, I saw, is that 3000 * 20/9 is the work done to counter balance resistance. So it's not really the new force the engine is applying. That's what I thought of it, that's how I could make sense out of it.

    Now if you tell me that the engine can indeed generate more force only because of gravity, I think it changes the problem and I will have to rethink it. So what do you say?


  20. Apr 7, 2010 #19
    The engine generates more force because the driver feels the car slowing down, and presses the gas pedal further to compensate. If the hill is steep enough and the engine not strong enough, then the driver will press the pedal all the way and the car will still slow down.

    If the driver kept the gas depressed the same, then the car would produce approximately the same force, and thus slow down.
  21. Apr 7, 2010 #20
    First of all, Lsos, I'm actually saying that in ALL situations, the driver presses the pedal "to the metal," so we're talking about the engine operating at full power.

    Okay Paulo, as soon as you understand this, I think you'll be good. I am not claiming that the engine is "thinking" or "smart." This is just a phenomenon which is best explained mathematically.

    Right now, you think an engine should always produce the same force. I'm claiming it will only produce the same "full power." Actually Paulo, there are many examples in physics in which one object produces very different amounts of force. For example, we could set up a pulley system in which a little boy weighs 50 kg could hang all his weight from the rope (and thus exert a force of 500 N on the rope) and he could lift a cinder-block twice his weight! But the pulley system is NOT adding energy to the boy. The boy is providing all the energy.

    So, wait a minute! you might say. If the boy can pull a rope with 500N of force in one situation, why should be be able to pull one with 1000N of force in another situation? Well, you will find that in both situations, force will be different, but power will be the same!!! Why? And how is energy conserved? Well, suppose the boy, with his arms exerting 500 N of force on his rope, is pulling his rope at a constant speed of, say, half-a-meter per second. Then the cinder block, weighing 1000 N, will be rising at a constant speed of a quarter-meter per second.

    Now, this situation is analogous to the car situation. The boy is like the engine, and the cinder-block is like the car. We don't see the engine. It's inside the car. But like the boy, it will push (lift) the car (cinder-block) with different forces even though it's the same engine (boy).

    Try very hard to understand this analogy, it's crucial to understanding power. It shows you how force will vary, but power may remain constant.

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