# Physics

A constant net force of 406 N, up, is applied to a stone that weighs 33 N. The upward force is applied through a distance of 2.4 m, and the stone is then released. To what height, from the point of release, will the stone rise?

The acceleration a is given by
a = F/m = 12.30 m/s^2
The velocity when released is
V = sqrt (2aX) = 7.68 m/s
The height H that it reaches after that is given by
gH = V^2/2
H = V^2/(2g)

Concern: The answer that I figured was incorrect. I got H = 3.013 m. Should I have subtracted the original height from that answer?

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1. Net Work= change in PE

(406)2.4= mgh
h= 406*2.4/33/g * g= 406*2.4/33

Now the rise after release will be 2.4 m less than h calculated above.

I don't get anything near what you got.

Secondly, the formula for acceleration you present ignores gravity, and gravity is definitely slowing the acceleration upward. The rest of your work, starting at the gH=v^2/2 makes no sense at all to me.

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