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Can someone help me with an average force problem?

  1. Oct 11, 2004 #1
    What is the average force exterted by a shot-putter on a 7.0 kg shot if the shot is moved through a distance of 2.8m and released with a speed of 13 m/s?

    Can someone help me out?
     
  2. jcsd
  3. Oct 11, 2004 #2
    One way to solve this is with conservation of energy. Recall work W=Fs. That is force times distance. Also recall kinetic energy K = (1/2)mv^2.

    This enough to get started?
     
  4. Oct 11, 2004 #3
    another approach would be to figure out the average acceleration of the shot using v^2=v(initial)^2+2ad. from there finding the force is fairly simple.
     
    Last edited: Oct 11, 2004
  5. Oct 11, 2004 #4

    BobG

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    You've got two equations: one for how far an object moves over a given amount of time, one for how fast an accelerating object moves over a given amount of time.

    For distance:

    [tex]s_f=s_i + v t + \frac{1}{2}at^2[/tex]
    [tex]2.8m=0 + 0 + \frac{1}{2}at^2[/tex]
    [tex]at*\frac{1}{2}t=2.8m[/tex]

    For velocity:

    [tex]v_f=v_i+at[/tex]
    [tex]13 m/s = 0 + at[/tex]
    [tex]at=13m/s[/tex]

    You can substitute 13m/s for at in the first equation and go from there.
     
  6. Oct 11, 2004 #5
    bobg: noticed a slight problem with that method. would need to get a third formula for the time because of the (at)*(.5)(t). you're able to substitute out the at, but there's still a t in there that you would need to deal with.

    the reason i chose the formula i did was because the only missing value in it was the acceleration, which happens to be the missing value from the righthand side of F=ma. a little algebra gives you a=v^2/(2d)
     
  7. Oct 12, 2004 #6

    BobG

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    Your way is easier and more efficient. I just thought the other would be easier to visualize what is happening.
     
  8. Oct 12, 2004 #7
    understood.
     
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