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Work an instantaneous power

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    An initially stationary 2.0kg object accelerates horizontally and uniformly to a speed of 10m/s in 3.0s. (a) In that 3.0s interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?


    2. Relevant equations
    (a)
    W= F*d
    a= (V-Vi)/t
    F=ma
    (x-xi)=Vt-(1/2)a(t^2)

    (b)
    p= F*V

    3. The attempt at a solution
    (a)
    I used the constant acceleration equation a= (V-Vi)/t and found that a= 3.33m/s^2
    then I found force, F=(2)(3.33)= 6.66N
    then I found the distance= (10m/s)(3s)-(1/2)(3.33)(3^2)= 15.015m
    W= (6.66N)(15.015m)= 100J
    I don't know that my logic was correct in finding work. I tend to overcomplicate things & i may not have found acceleration correctly.
    (b)
    for instantaneous power i used p= F*V (dot product) so Fx*Vx= (6.66N)(10m/s)= 66.6W? I assumed my velocity at the end of the interval was 10m/s. And I assumed my force component would be the same because the problem is one dimensional.
    (c)
    t=1.5s so I plugged that time in to find a= 6.66m/s^2 and F=13.32N
    Then p=(13.32N)(10m/s)= 133.2?? Am I supposed to find V when d= 7.5m?

    I'm doing review problems for my midterm and I don't have access to a lot of the solutions. Any help would be appreciated. I just want to know if I'm doing these problems right & if not, then what's wrong with my logic. Thanks again.
     
  2. jcsd
  3. Nov 14, 2009 #2

    kuruman

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    Science Advisor
    Homework Helper
    Gold Member

    Parts (a) and (b) are correct.

    For part (c) the instantaneous power is indeed Fv, but the acceleration is constant. You cannot recalculate it. Just find v at 1.5 s and then find Fv. The power is "instantaneous" because the velocity is instantaneous.
     
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