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Homework Help: Sprinter's Power Output

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    A 48.0 kg sprinter, starting from rest, runs 46.0 m in 7.30 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 1.70 s?

    2. Relevant equations
    [tex]x=x_0 + v_0 t + \frac{1}{2}at^2[/tex]
    [tex]\vec{F}=ma[/tex]
    [tex]v=v_0 + at[/tex]
    [tex]KE=\frac{1}{2}mv^2[/tex]
    [tex]P=\frac{\Delta E}{\Delta t}[/tex]

    3. The attempt at a solution
    For part A, I first used [tex]x=x_0 + v_0 t + \frac{1}{2}at^2[/tex] to find the acceleration of the sprinter:

    [tex]x=x_0 + v_0 t + \frac{1}{2}at^2[/tex]
    [tex]46=0 + 0(7.3) + \frac{1}{2}a(7.3)^2[/tex]
    [tex]a=1.726 m/s^2[/tex]

    With the acceleration, I stuck it into Newton's Second Law and found the force.

    [tex]\vec{F}=ma[/tex]
    [tex]\vec{F}=48(1.726)[/tex]
    [tex]\vec{F}=82.9 N[/tex]

    My answer for part A was correct.

    Part B is where I am having some difficulties. First I got the velocity of the runner at [tex]1.7 sec[/tex]:

    [tex]v=v_0 + at[/tex]
    [tex]v=0 + (1.726)(1.7)[/tex]
    [tex]v=2.934 m/s[/tex]

    I then calculated the amount of work done by calculating the kinetic energy, which I used because the sprinter is running and in motion.

    [tex]KE=\frac{1}{2}mv^2[/tex]
    [tex]KE=\frac{1}{2}(48)(2.934)^2[/tex]
    [tex]KE=206.6 J[/tex]

    Then to calculate power, I took the change in the kinetic energy (0 J to 206.6 J) and divided it by the change in time.

    [tex]P=\frac{\Delta E}{\Delta t}[/tex]
    [tex]P=\frac{206.6}{1.7}[/tex]
    [tex]P=121.55 W[/tex]

    That answer was marked incorrect. I then tried to add the horizontal force (the answer to part A), which added up to 204.45 W and that was also incorrect.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. May 4, 2010 #2
    Nevermind, I figured it out.

    I did P=F*v which gave me the correct answer.
     
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