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Calculate F given m, d, and v should be easy.

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the average force exerted by a shot putter on a 7.0-kg shot if the shot is moved through a distance of 2.8m and is released with a speed of 13m/s?



    2. Relevant equations
    F = m * a
    a = Δv/t
    v = Δd/t


    3. The attempt at a solution

    This is a problem that should be easy, I think. I get one answer, but the answer in the back of the book for this particular problem is way different.

    t = d/v = 2.8m/13m/s = 0.22 seconds

    a = v/t = 13m/s / 0.22s = 59.09 m/s/s

    F = ma = 7kg * 59.09 m/s/s = 413.6 N....right?

    My calculation of 413.6N doesn't agree with the book's 210N in the answer key....
     
  2. jcsd
  3. Jun 12, 2012 #2
    You have to use average velocity not final velocity.
    Or just constant acceleration(average force) with final velocity, 13m/s
     
    Last edited: Jun 12, 2012
  4. Jun 12, 2012 #3
    Are you familiar with work and kinetic energy? If so, setting the kinetic energy [itex]mv^2/2[/itex] equal to the work done by the force [itex]Fd[/itex] would be an appropriate method.
     
  5. Jun 12, 2012 #4
    That makes sense.... so my answer given above would be the instantaneous force right at the release rather than the average throughout the whole thing?
     
  6. Jun 12, 2012 #5
    I guess it's not.
    What we are given the distance travel from 0 to final velocity.
    If we apply constant force(acceleration), we will find the time taken to cover that distance.
    The actual force applied is not linear.

    Average velocity x time=distance
     
    Last edited: Jun 12, 2012
  7. Jun 14, 2012 #6
    final kin energy is E=0.5m*v^2, where m=7kg, v=13m/s;
    work done on the shot is E=F*s, where s=2.8m, F is force by putter;
    energy conserves; E=E; 0.5m*v^2= F*s, hence
    F=0.5m*v^2 / s =
     
  8. Jun 14, 2012 #7

    Curious3141

    User Avatar
    Homework Helper

    Mathematically, the average force is given by:

    [tex]\bar{F} = \frac{1}{s}\int_0^s Fdx[/tex]

    The definite integral is work done on the object, which is equal to the final kinetic energy, which is why you end up with:

    [tex]\bar{F} = (\frac{1}{s})(\frac{1}{2}mv^2)[/tex].
     
  9. Jun 14, 2012 #8
    Simplest way to solve this problem,

    Use work energy theorem,

    fcosθd=(0.5)mv^2 (cosθ= 1 assuming all force is in the same direction as the displacement)

    Rearrange algebraicly and you'll get the answer.

    Hope this helps!
     
  10. Jun 14, 2012 #9
    I thought the same!
     
  11. Jun 14, 2012 #10
    http://img864.imageshack.us/img864/4082/collision.jpg [Broken]

    http://img256.imageshack.us/img256/5251/lolljg.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
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