How Does Kinetic Energy Relate to Forces in Shot Put?

[jjbrinkman3]

During a throw a shot is initially swung around in a circle reaching a speed of about 3.5 m/s?
during a throw a shot is initially swung around in a circle reaching a speed of about 3.5 m/s. it is then accelerated more or less in a straight line over a distance of about 1.7 m, leaving the hand at roughly 14 m/s. The shot is 7.26 kg.

what is the average force exerted on the shot during this latter part of the launch? compare this to the measured peak force of 600 N. Is that reasonable?

I think you use the equation F=W/d where W=667 J and d=1.7 m
this gives you 392.4 N.
Did I do this correct?
How does this compare to the peak of 600 N and is this reasonable?

 

[PeterO]

During a throw a shot is initially swung around in a circle reaching a speed of about 3.5 m/s?
during a throw a shot is initially swung around in a circle reaching a speed of about 3.5 m/s. it is then accelerated more or less in a straight line over a distance of about 1.7 m, leaving the hand at roughly 14 m/s. The shot is 7.26 kg.

what is the average force exerted on the shot during this latter part of the launch? compare this to the measured peak force of 600 N. Is that reasonable?

I think you use the equation F=W/d where W=667 J and d=1.7 m
this gives you 392.4 N.
Did I do this correct?
How does this compare to the peak of 600 N and is this reasonable?

Why do you think W = 667 J? Not saying you are wrong, just wanting to see how you calculated that.

 

[SammyS]

Welcome to PH.

That all looks reasonable.

 

[jjbrinkman3]

Why do you think W = 667 J? Not saying you are wrong, just wanting to see how you calculated that.

I found the KE in the first part by KE=1/2mV²
this gives you 44.4675 J.

Then i found the KE of part 2 using the same equation which is 711.48 J.

Then I found ΔKE=KE_f-KE_i which gives you 667 J.

 

[PeterO]

I found the KE in the first part by KE=1/2mV²
this gives you 44.4675 J.

Then i found the KE of part 2 using the same equation which is 711.48 J.

Then I found ΔKE=KE_f-KE_i which gives you 667 J.

That's fine, just wanted to check you were calculating correctly.

I think the question is drawing yo to the idea that the thrower [putter] will not supply a constant force, but one which increases from zero to a peak, then drops back to zero. That will partly be affected by the angle of your arm, and ability to apply force with your elbow bent by various amounts.
If the force increased uniformly, then reduced uniformly, the peak force would be double the average. How does that compare here?

 

[jjbrinkman3]

That's fine, just wanted to check you were calculating correctly.

I think the question is drawing yo to the idea that the thrower [putter] will not supply a constant force, but one which increases from zero to a peak, then drops back to zero. That will partly be affected by the angle of your arm, and ability to apply force with your elbow bent by various amounts.
If the force increased uniformly, then reduced uniformly, the peak force would be double the average. How does that compare here?

This is almost 200 N more than the peak of 600 N, so this answer is not reasonable. Does that make sense?

 

[PeterO]

This is almost 200 N more than the peak of 600 N, so this answer is not reasonable. Does that make sense?

Not really, a human may be able to increase the force quickly and maintain his peak effort for a time before dropping off. I am not an exercise physiologist, so I don't really know how the human body can exert a force.
I do know that if the object you are throwing gains sufficient speed I can no longer apply a force, since it is already moving as fast as I can move my hand.