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!

Calculating Power

  1. May 9, 2013 #1
    An Aerodynamic (friction-less) car takes about 270 Newtons of force to maintain a speed of 25m/s.



    How much horsepower is required from the engine to maintain that speed?
    How much horsepower is required for the same car to accelerate from 0-25m/s in 6 seconds?




    I thought it was a pretty easy problem then I realized I couldn't divide joules by any sort of time.
    Ek=(1/2)(1000kg)(25m/s)^2 which gets me 25,000J, but power is J/t and I only have J, I somehow need to figure out how much time it takes to maintain 25 m/s from a force of 270N or have time in somewhere and I tried saying
    270N = 1000kg * 25m/s * (1/t), which gave me 270s when I solved for t. Is that a coincidence or is that some mathematical tautology that tells me nothing?
    It says 25m/s, can I just assume I divide by 1 second? Or what? How do I get a time from this to calculate power?
    Thanks.
     
  2. jcsd
  3. May 9, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    The definition of power is work done per unit time. You are give that it takes 270 N to maintain a speed of 25 m/s. Work is defined as force times distance. In 1 sec, how much work is done by the car's engine in maintaining this speed?
     
  4. May 9, 2013 #3
    It seems like you are using the term "work" in place of energy which I'm not familiar with yet. Energy is force acted over a distance, or N * m, but so is work according to you, but in this context it doesn't make sense to say energy was gained though I suppose it would make sense if chemical energy was being converted to kinetic energy, but if it was friction less you would only need to apply the energy once for the car to keep going and it should be accelerating if you keep giving the train car energy...not sure what's going on here. In any case, the answer to your question seems to be 1350J or whatever the units of work are. Not sure why exactly its a different answer than what I got for kinetic energy.
     
  5. May 9, 2013 #4
    Ok well I got a new hypothesis but I don't know if it will work, I looked at the units some more, and it seems likely that I can do force * velocity, I'll have 1/t^3 and a m^2. Is that right?
     
  6. May 9, 2013 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, force has units. In the SI system, force is a derived quantity. Ditto for joules, too.

    I'm not sure what you definition of work is, but W = F * d is pretty basic physics:
    http://en.wikipedia.org/wiki/Work_(physics)
     
  7. May 9, 2013 #6

    gneill

    User Avatar

    Staff: Mentor

    Is the question complete? Without the mass of the car how can we determine the amount of kinetic energy that has to be invested in order to bring it up to speed? (Maintaining the speed is another matter, since we are given the force required to counteract any losses and the velocity that is being maintained).
     
  8. May 9, 2013 #7
    Forgot to mention the car is 1000kg, but maybe it could have been derived from the Newtons. Anyway, isn't Force times Distance also the units for Joules? I could have sworn it was, kgm^2/s^2, the force acted over a certain distance. That would make even more sense when calculating power from force*velocity because joules/time = kgm*2/s^3 whic his only a 1/s difference from your definition of work.
     
  9. May 9, 2013 #8

    gneill

    User Avatar

    Staff: Mentor

    It makes a big difference! The force required to maintain a given velocity only has to overcome losses (friction). On the other hand, changing the velocity (as in acceleration) requires an energy investment in the kinetic energy of the body, which depends upon the mass, as well as overcoming frictional losses.

    Yes, f*d yields work. And f*v yields power (in this case constant power required to offset frictional losses at a given constant velocity).
     
    Last edited: May 9, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculating Power
  1. Power calculation (Replies: 1)

  2. Calculate the power (Replies: 2)

  3. Power calculation (Replies: 1)

  4. Power Calculations (Replies: 10)

  5. Calculating power! (Replies: 3)

Loading...