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Constant power? Instant Power?

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Starting from rest at t = 0, a 5.0-kg block is pulled across a horizontal surface by a constant horizontal force having a magnitude of 12 N. If the coefficient of friction between the block and the surface is 0.20, at what rate is the 12-N force doing work at t = 5.0 s?

    2. Relevant equations

    Force = mass x acceleration
    Friction force = coeff of friction x normal force
    Work = mass x acceleration x distance
    Power = mass x acceleration x distance / time ... ?

    3. The attempt at a solution

    Force in the horizontal direction should be applied force minus friction force right? So:

    12.0 newtons - (0.20 x 5.0 kg x 9.8) newtons = 2.20 newtons
    acceleration in x direction = 2.2 newtons / 5 kg = .44 meters per second squared
    final velocity = acceleration x time = .44 x 5 = 2.2 m/s
    power = force x speed = 2.2 newtons x 2.2 meters per second = 4.84 ?
    or... mass x acceleration x distance = work per second? That would be:
    5 kg x .44 m/s^2 x 5.5 m / 1 s = 12.1 watts ?

    0.13 kW
    0.14 kW
    0.12 kW
    26 W
    12 W

    What I don't understand is what "power at time x" means. Wouldn't that be instantaneous power? I thought power was total work over a time interval. I'm confused as hell ><
     
  2. jcsd
  3. Oct 24, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Work is F * D and Power is W / time.

    Rewriting then Power = F * D/time

    If you want instantaneous power then choose a ΔD and divide by the corresponding Δt.

    Power = F * Δd/Δt ... hmmm... anything look familiar?
     
  4. Nov 3, 2008 #3
    So for that specific instant in time, the power is force in that direction multiplied by speed at that point?
     
  5. Nov 3, 2008 #4
    Ok THIS IS DRIVING ME INSANE ALSDKJFA;LSDKJFA;LSKDJF

    Here is what I keep getting every time I work this. EVERY.... TIME....

    Starting from rest at t = 0, a 5.0-kg block is pulled across a horizontal surface by a constant horizontal force having a magnitude of 12 N. If the coefficient of friction between the block and the surface is 0.20, at what rate is the 12-N force doing work at t = 5.0 s?

    The force in that direction of motion is 2.2 newtons ---- 12 N - (.2x5x9.8) = 2.2 newtons

    The acceleration --- 2.2 N = (5 kg) x (a) .... so a = 0.44 meters per second squared.

    Since force is constant, acceleration is constant. Starting from rest:

    velocity at 5 seconds is (0.44 m/s^2)(5) = 2.2 meters per second. HEY LOOK AT THAT!!!! THE SPEED IS THE SAME AS THE FORCE. HUR HUR. WTFFFF?????????

    So if power is the rate of work, and somehow instantaneous work is force times velocity at a point, then ------- 2.2 newtons TIMES velocity at t=5 (ALSO 2.2 HURHUR WTF) = 4.4 watts.

    But guess what? 4.4 watts isn't one of the choices. AWESOME. COOL QUESTION.a s;ldkfja;slkdjfal;skdjfldffffffffffffff

    H E L P
     
  6. Nov 3, 2008 #5

    LowlyPion

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    Homework Helper

    OK. You figured a = .44

    P = F * D / T

    You know Force is 2.2 N

    Your Distance is D = 1/2 * a * T2 = 1/2 * .44 * 52

    P = 2.2 * .22 * 52 / 5 = 2.2 * .22 * 5 = 2.42 N-m/s
     
  7. Nov 4, 2008 #6
    Thank you for the help, and for putting up with my frustration, but 2.42 newton meters per second aka watts is not one of the listed solutions.

    Could it be that I should use the 12 newtons of applied force x 2.2 m/s : 26.4 watts? That's one of the solutions given.

    I think this question is so disgustingly hard for the simple fact that the speed also just happens to be equal to the net force on the object.

    Could 26.4 watts be a possible solution?
     
  8. Nov 4, 2008 #7

    LowlyPion

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    I misread. The work from the 12 N force over the distance is what you want.

    That would be 12 * 1/2*a*t2/t = 12*1/2*.44*5 = 13.2 N-m/s
     
  9. Nov 4, 2008 #8
    Not sure, but isn't 13.2 newton meters per second = 13.2 watts?

    If so, the solution set for the question still doesn't provide a match: closest is .13 kW, which would be 130 watts.

    They ask what rate the 12 newton force is doing work at time = 5. I've look around the web and found a few sources that say power at a moment in time is the force times the change in distance over change in time (aka speed).

    Given the acceleration is .44 meters per second squared, and that initial velocity is zero, speed at time = 5 should be 2.2 meters per second.

    So if power at an instant is force times speed, answer would be 26.4 watts, no?
     
  10. Nov 4, 2008 #9
    Alright my instructor gave me the answer on this one (for anyone who arrives here via google):

    You have to
    1. Calculate net force as the difference between F and (Uk)mg = 2.2 newtons.
    2. Use that to find acceleration using F=ma
    3. Use the acceleration to find the speed at time = 5 seconds.
    4. Multiply that speed with the 12 newton force to get instantaneous power at time = 5 seconds. ->26.4 watts

    This question sucked because the force of kinetic friction just happened to be the exact same number as acceleration due to gravity (9.8) AND the net force was equal to the velocity at time = 5 seconds.
     
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