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Force Problem with football punter

  1. Sep 21, 2007 #1
    [SOLVED] Force Problem

    1. The problem statement, all variables and given/known data

    2. A football punter accelerates a football from rest to a speed of 10m/s during the time in which his toe is in contact with the ball (about 0.20sec). If the football has a mass of 0.50kg, what average force does the punter exert on the ball?

    2. Relevant equations
    so far i have this and can't seem to find the equation(s) but i got some and does not correspond with the correct answer.

    F= thats what we need for answer

    so F=ma

    3. The attempt at a solution
    but i dont understand how we put our time in an equation to solve for Force, I know im missing something but i don't know what. I have the answer, answer only which does not help at all :P. Thats to check whether we get the correct problem

    F = ma
    F = 10m/s (.50kg)
    F = 5N (?) but answer that was given was 25 N

    I want to know how to do it :P even thou teacher gives us answer for homework. Which is the most important thing
    I'm stuck with this help

    Thank you in advance
  2. jcsd
  3. Sep 21, 2007 #2


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    Science Advisor

    F = 10m/s (.50kg) is velocity * mass or mv which is the final momentum.

    F = dp/dt or m dv/dt so take mass * (change in v)/(change in time)

    a = dv/dt
  4. Sep 21, 2007 #3
    so F=ma:

    F= 10ms/(.50kg) = 5 m/s correct which equals Velocity(?)

    so for F= dp/dt what is dp? change in P(?) / change in time so we get

    F= 5ms/.20s = 25N

    my question comes from F=dp/dt we use that equation to find out Force? Where can we say that F=dp/dt (as u used) comes from?

    srry if i sound confusing
  5. Sep 21, 2007 #4


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    Science Advisor

    p = m v = momentum.

    Now let's take the case for constant mass.

    F = dp/dt = d (mv) /dt = m dv/dt = m a

    Back the problem.

    m = 0.5 kg, which is accelerated from rest (v=0) such that at the end of 0.2 s, the mass has a velocity 10 m/s. So the change in velocity (dv) = 10 m/s - 0 m/s, and the change it time (or time interval), dt = 0.2 s.

    so F = ma = m dv/dt

    m = 0.5 kg, dv = 10 m/s, and dt = 0.2 s
  6. Sep 23, 2007 #5
    got it :p that helped out
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