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Average and instantaneous power-please help

  1. Feb 19, 2006 #1

    1.Power is defined as W / t. But problem is that the work done by the force increases with displacement, since body travels same distance in ever shorter time periods. So how can we find instantaneus power if the more time it passes by more work does force do in same time interval ?


    P = W/t = F * s/t = F * s/t = F * v

    I understand how we derived this formula but have no clue why it's legit.
    For one thing v is average speed over distance s, so why can we use it in formula to get power?

    I'm sorry for not being able to articulate my question any better(),
    but I do hope someone will be able to figure out what I'm trying to ask and explain it .

    3.Is P = F * v instantaneus power? If so, why would this formula give us instantaneus power?

    4.Most machines are designed and built to do work on objects. These machines are described by a power rating. The power rating indicates the rate at which that machine can do work upon other objects. Thus, the power of a machine is the work/time ratio for that particular machine. Over what period of time is work measured for that machine?
    Even if machine has power rating of 10kW/1 day, wouldn't in two days of non-stop running have even greater power?
    I'm also aware that this can't go on for ever, since rotor or whatever at some point reaches its final speed and thus can no longer accelerate and have its kinetic energy changed. What happens then? Is at that moment power=zero?

    thank you
  2. jcsd
  3. Feb 19, 2006 #2


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    Power will be a function of t. Essentially, you form the "average" power fracation and take the limit as the time interval goes to 0:

    As long as v is constant, average speed is the same as "instantaneous" speed. This formula works for F and v constant. If either or both are not constant, then, again, it becomes a derivative:
    [tex]W= \int F(x)dx[/tex]
    [tex]P= \frac{dW}{dt}= \frac{dW}{dx}\frac{dx}{dt}= F(x(t))v[/tex]

    Power measures RATE at which work is done. No, running for two days at full power, a machine would do more WORK but it's power is a constant.

    Driving at 60 miles per hour for two hours, you would go farther than in one hour, but your speed would remain the same.
  4. Feb 20, 2006 #3
    I know very little about derivatives but couldn't we find instantaneus power at specific time T by finding derivative of a function work(T)?

    But if v is constant then no work is being performed and so by definition formula [tex]\frac{A}{t } = F\frac{s}{t}[/tex] shouldn't be valid!!!

    I don't think you understood my question. English isn't my first language so please have patience. We have a machine, a rotor of some kind that is spinning. That spinning rotor is doing work on objects it comes in contact with (perhaps it cuts grass - if that is the case, then it does work on grass).
    Anyhow, for rotor to be doing any kind of work there must be same work (friction ignored) performed on rotor itself in order for rotor to be spinning.

    BTW-I know formula for power when object is spinning is
    [tex]P = torque*\frac{angle}{time}[/tex] ,

    but since I haven't started learning that yet and since I'm interested in learning about power in general, I will allow myself some artistic freedom and pretend formula for power when object is roating is

    [tex]P = F\frac{s}{t}[/tex] .

    At time [tex]T1=0 AM[/tex] we apply force F on some point on rotor. So if this point at time [tex]T1=1 AM [/tex] travels (due to constant force F) distance [tex]s1=1000 km[/tex], we say it has [tex]power = F*s1/(T1-T0)[/tex] . Due to force being constant speed of rotor gets ever faster and in the first hour the traveled distance was 1000 km. But from time T1 to time T2 this same point will travel even greater distance and thus do more work in same time period due to its velocity being greater.
    From time [tex]T1=1 AM[/tex] to time [tex]T2=2 AM[/tex] point on rotor travels distance [tex]s2[/tex] , where [tex]s2 - s1 > s1 [/tex] and

    [tex] F * \frac{(s2-s1)}{(T2-T1)} > F * \frac{s1}{(T1-T0)}[/tex]

    You see what's bothering me?
    Last edited: Feb 20, 2006
  5. Feb 20, 2006 #4


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    That isn't true. If v is constant, it only means the force is balanced by another force, like friction. But work is definitely being done.
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