Force Exerted by Shot-Putter: Solving the Puzzle

[TheNovice]

1. What is the average force exerted by a shot-putter on a 7.0 kilogram shot if the shot is moved through a distance of 2.8 meters and is released with a speed of 13 m/s?



2. F=ma, x = vo t + 1/2 at^2, v=vo at



3. I think you get time by using D = vt, but when I use this time and plug it into the distance equation above I get a weird answer. The real answer should be in the ballpark of 200 Newtons.

 

[Heimisson]

ok I'm not sure how much calculus you know but the chain rule allows you to write:
F=ma=m\frac{dv}{dx}\frac{dx}{dt}=m\frac{dv}{dx}v
this means the F =mv*(change in velocity)/(change in position)


However this gives me about 400 N so I might be forgetting something.

 

[TheNovice]

ok I'm not sure how much calculus you know but the chain rule allows you to write:
F=ma=m\frac{dv}{dx}\frac{dx}{dt}=m\frac{dv}{dx}v
this means the F =mv*(change in velocity)/(change in position)


However this gives me about 400 N so I might be forgetting something.

Thanks for the response

But I'm in an introductory physics course; calculus is a bit too much for me.

I'll play around with the numbers some more.

The problem seems really easy... I just can't get the right answer.