Shot-putter

[kbrowne29]

I'm having difficulty with the following problem: 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 2.8 m and is released with a speed of 13 m/s.

I know that I have to use the formula F=ma. The mass is obviously 7.0 kg, but I don't know how to find the horizontal component of the acceleration of the shot. I don't know how to incorporate the initial vellocity and the distance the shot travels. I would appreciate any help. Thanks.

[jcsd]

Use the formula:

v2 = u2 + 2as

[kbrowne29]

what do the variables stand for?

[jcsd]

v is final velocity (13ms-1), u is initial velocity (assume this to be zero), a is accleration, s is displacement (distance).

[kbrowne29]

ok, so that gives an acceleration of 2.32m/s^2. But it doesn't make sense that this would be the acceleration to plug into the formula F=ma. I know that the correct answer to the problem is 210 N.

[jcsd]

Show me your workings, using the above equations I got an answer of 211.25 N (210 N if I'd rounded the accleration to 3 ms-2).

[kbrowne29]

I got it! I've been making such a dumb mistake. When I used v^2=u^2 + 2as, I kept forgetting to square the 13! Now i'm getting the right answer. Thanks a lot for helping me out here.