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Homework Help: Sprinter Work and Power

  1. Jun 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A 50 kg sprinter, starting from rest, runs 50 m in 7.0 s at constant acceleration.
    a. What is the magnitude of the horizontal force acting on the sprinter?
    b. What is the sprinter's power output at 2.0 s, 4.0 s and 6.0 s?

    2. Relevant equations

    xf= xo + vt+ 0.5*a*t^2
    F= ma

    vf= vi + at
    P= (delta Work)/(Delta t)
    P= f*d*cos angle

    3. The attempt at a solution

    Using the distance, initial velocity, and time I solve for acceleration

    xf= xo + vt+ 0.5*a*t^2
    50 = 0 + 0+ 0.5*a*(7^2)
    a= 2.04 m/s^2

    Therefore, F=ma=50 (2.04)=102 N

    This, I got right. But when I calculate the power, I'm not getting the right answer when i use the first equation.
    I calculate the velocity at t= 2s, and I get
    vf = vi + at
    vf= 2.04 ( 2)
    vf= 4.08 m/s

    Therefore work=delta kinetic energy= 0.5 m (vf^2 - vi^2)
    K= 0.5 ( 50) ( 4.08^2)
    K= 416.16 J

    Therefore delta work=delta kinetic energy=416.16 J

    Therefore Power= 416.16/ 2 s = 208.08 W.

    Apparently the answer is just 416.16 J. When I use the second equation of P= F*v=102*4.08=416 J, I get the correct answer. Why are these two equations giving me different answers?
  2. jcsd
  3. Jun 18, 2008 #2


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    Science Advisor
    Homework Helper

    Taking the difference in the kinetic energies at t=0 and t=2 and dividing by 2 sec doesn't work for the same reason that taking the difference in the positions at t=0 and t=2 and dividing by 2 sec doesn't give you the velocity at t=0. They want an 'instantaneous power' not an 'average power'.
  4. Jun 18, 2008 #3
    I didn't know that there was a difference between the two. So whenever I have a velocity I should use the second one?
  5. Jun 18, 2008 #4


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    Science Advisor
    Homework Helper

    There's definitely a difference between average and instantaneous. If they don't ask for average then they probably mean instantaneous.
  6. Jun 18, 2008 #5
    Thanks a lot! I understand it now =).
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