1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need a better understanding of the formula P= F v

  1. Feb 21, 2012 #1
    Hello Forum,

    Work is given by force times distance: W= F d. Power P is work over time: W/t or F v. If the force is constant the velocity is changing with time: v=v(t).

    So instantaneous power is P_inst= F(t) v(t).

    Can we have a situation where v is constant and F is constant too, i.e. the instantaneous power is constant? That seems strange, since if the force is constant the velocity must automatically be changing with time t....

    thanks
    fisico30
     
  2. jcsd
  3. Feb 21, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    An applied force delivering power is not necessarily the only force or the net force acting on the system.
     
  4. Feb 21, 2012 #3
    The average power would be:

    P_avg= Integral of P(t) dt

    Or P_avg= total Work/ time

    If a force is a function of space, F(x), it changes from point to point. But is it also a function of time, since from the perspective of the object on which the force acts, the force appears to change with time too.....
     
  5. Feb 21, 2012 #4
    Sure Doc Al,

    but if an object is moving at constant speed, there is not net force on it.

    So, in the formula P=F v, if v=constant, F should be zero....

    For instance, a car travelling at constant speed on the road: the engine provides a force, and air drag+ friction provide an equal and opposite force. Net force is zero. Net work is zero.
    Net power zero too...

    I guess the formula P= F v would give the power that either the engine (or the total resistance force) would give....
     
  6. Feb 21, 2012 #5

    Pengwuino

    User Avatar
    Gold Member

    Sure. Consider a block being pushed along a surface with friction. Since the kinetic frictional force is given by [itex]F_{\mu} = \mu mg[/itex], where [itex]\mu[/itex] is the coefficient of kinetic friction, if you provided that same force to push the block, you would have no acceleration, a constant velocity, and a non-zero force being applied by you. Your power output would be constant.

    The problem that you're having seems to be the fact that you're looking at the whole system when you need to only look at parts of the system. Since energy is conserved for the whole system, you will always find the total power in a system is 0, but that's only if you look at the whole system. In my example, you would want to know how much power it would take to push the block, but almost never have any interest in how that power gets transferred into the surface. The same goes for a car. We're very interested in the power output by an engine, but we don't care how that power gets transferred to the air, the ground, thermal energy, etc.
     
    Last edited: Feb 21, 2012
  7. Feb 21, 2012 #6
    I guess this will sound kind of silly but the case with constant force velocity and power is when total force is zero, velocity is constant and total power is zero too. :D.

    But for example there might be two equal and opposite forces that keep the velocity constant, the total power will be zero but the power of each force will not be zero but will be equal and opposite (so they addup to zero total power).
     
  8. Feb 21, 2012 #7

    Doc Al

    User Avatar

    Staff: Mentor

    True.
    No. Only the net force is zero, not the applied force that you might be interested in.
     
  9. Feb 21, 2012 #8
    I totally agree Doc Al, but somehow the formula is misleading to me:

    P= F v. the force F is one of the forces, but the velocity v is not the v caused by that specific force. It is the velocity caused by the net force...
    somehow the formula seems to mix what a single force does with the velocity that the net force causes....
     
  10. Feb 21, 2012 #9

    Pengwuino

    User Avatar
    Gold Member

    There's nothing wrong with that. Consider kinematic equations. Consider a cannon shooting an object and the kinematic equations include [itex]x(t) = x_0 + v_0t + {{1}\over{2}}gt^2[/itex]. The acceleration is caused by gravity, but the initial velocity was given the cannon. It doesn't matter what causes certain parameters of a system to be what they are, all that matters is that they are what they are. That's actually one of the beautiful things about physics.
     
  11. Feb 21, 2012 #10

    Doc Al

    User Avatar

    Staff: Mentor

    It doesn't matter what 'causes' the velocity, just that the applied force is exerted through a distance at a certain rate. (And better to think of forces as causing changes in velocity, not velocity.)
     
  12. Mar 20, 2012 #11
    Hello Doc Al,
    consider the following situation: a block of mass m=1 Kg sliding on a surface with coefficient of kinetic friction mu=0.3

    The block is moving at constant speed v1=2 m/s. That means the input force F is equal to the kinetic friction f_k.
    We want to bring the block to a new constant velocity v2=4 m/s: F must become larger than the kinetic friction for a certain period of time, until the speed becomes 4 m/s. At that point the force F can return to be equal to f_k to maintain constant speed.

    Now the power P= F v. F is the same when v1 and v2, but the power seems to be different:

    P1= F v1 and P2= F v2. but the force F should be putting the same amount of power in the block once the constant velocity is reached since F is the same in both situations...

    fisico30
     
  13. Mar 20, 2012 #12

    Doc Al

    User Avatar

    Staff: Mentor

    Why do you think the power will be the same when the speeds are different? Think about it: The work done (against friction, in this case) is Force*distance. Since more distance is covered per unit time with a faster speed, the power must be greater--even though the force is the same.
     
  14. Mar 20, 2012 #13

    rcgldr

    User Avatar
    Homework Helper

    Only if there is an opposing force, such as gravity, lke driving a car uphill. There is no net force on the car, but in this case, assuming no losses to other forces (aerodynamic drag, rolling resistance, ...) power x time ends up as increase in gravitational potential energy instead of kinetic energy.
     
  15. Mar 22, 2012 #14
    Doc Al,

    you say: "Since more distance is covered per unit time with a faster speed, the power must be greater--even though the force is the same."

    I agree within the unit time more distance is covered since the speed is higher. Hence more work. It was hard to understand that the energy input source, still exerting the same force F, would be able to input more energy.
    If the time interval is 1 second and v1=10 m/s and v2=20 m/s, the force F=f_k=2 N.

    The power that we need to input is 2*10=20 J/s and 2*20=40 J/s. Even if the force is the same and it t is being applied for the same amount of time (1 second), the energy input is larger because the covered distance is larger.

    What sound weird is that we have the same force F for the same amount of time. That would lead me to conclude that this force F is delivering the same amount of energy to the block. But it is not.......

    fisico30
     
  16. Mar 22, 2012 #15

    rcgldr

    User Avatar
    Homework Helper

    Force x time is called impulse, and it results in a change in momentum. Assuming mass is constant, then

    m v1 = m v0 + F t

    v1 = v0 + (F / m) t

    if force is constant, then a = F/m

    v1 = v0 + a t

    (F/m) t = a t = (v1 - v0)

    kinetic energy = 1/2 m v 2

    work done or change in energy =

    ΔE = 1/2 m (v12 - v02)

    ΔE = 1/2 m (v1+v0) (v1-v0)

    ΔE = 1/2 m (v1+v0) a t

    since a is constant

    1/2 (v1+v0) = vavg (average velocity)

    ΔE = (m a) (vavg t)

    distance = average velocity x time: d = (vavg) t

    ΔE = (m a) d = F d

    power = ΔE / Δt = F d / Δt = F v

    For non-constant force and acceleration, you'd have to use calculus, but you'd still end up with P = F v.
     
    Last edited: Mar 22, 2012
  17. Mar 23, 2012 #16

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    ?? sounds [STRIKE]quirky[/STRIKE] implausible to me.. I can certainly think up things that aren't [STRIKE]so[/STRIKE] physically workable,, so let's see here.

    What would happen if the mass of the thing were to increase as f(time)

    f = ma
    v = v0 + at

    Just to make the point, Let m = kt

    a = f/m = f/kt

    so, v = v0 + (f/kt) X t

    gives v = v0 + f/k and those are all constants.


    Do things in motion become more massive? I guess that could be why clocks in motion slow down... /wiseguy icon/

    Anyone else having trouble with latex tonite? Seems to bomb my iexplorer.
     
    Last edited: Mar 23, 2012
  18. Jul 7, 2012 #17
    In summary:

    in the power expression P=F v, v is the instantaneous velocity of the object and F is not the net force on the object, correct?

    If F was the net force (due to whatever multiple forces are acting ON the object) the speed would be necessarily changing and be a function of time t.

    If v(t)=constant, then F must be one of the forces on the object that is delivering or subtracting power to the object.

    It is possible that P(t)=F(t) v(t). This expression would indicate the instantaneous power. The object velocity changes with time and F(t) is a one of the (time-changing) forces on the object.

    What if v(t)=constant and F=F(t), i.e. it changes with time? The power delivered or subtracted by the force F(t) is not constant in time. The object moves at constant velocity because there are other forces ON the object that are also changing with time, balancing out the delivery of subtraction of power by the force F(t), so that v remains constant and no net power is added to the object, correct?

    If net power is added (or subtracted from) to the object, then v must necessarily change....

    thanks,
    fisico30
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Need a better understanding of the formula P= F v
Loading...