How Is Average Force Calculated in Shot Put?

[Sima]

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?

 

[Soveraign]

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?

 

[Vadim]

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.

 

[BobG]

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:

$$s_f=s_i + v t + \frac{1}{2}at^2$$
$$2.8m=0 + 0 + \frac{1}{2}at^2$$
$$at*\frac{1}{2}t=2.8m$$

For velocity:

$$v_f=v_i+at$$
$$13 m/s = 0 + at$$
$$at=13m/s$$

You can substitute 13m/s for at in the first equation and go from there.

 

[Vadim]

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)

 

[BobG]

Your way is easier and more efficient. I just thought the other would be easier to visualize what is happening.

 

[Vadim]

understood.