Quote:
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Originally Posted by KenWittlief
using F=M*A and V=A*t my calculations indicate you only need 19 lbs of force over a distance of 6" to accelerate a 0.175 kg ball up to 12m/s
the acceleration is 480M/S^2, and the time to get from 0 to 12m/S is 25mS
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There's something wrong here. All the equations make sense. However, the solutions that use said equations don't. Actually...nevermind, in the middle of writing this post I found the descrepency.
Ken's solution:
F=ma => a=F/m
19lb = 84.5N
a= 84.5/0.175 = 480m/s^2
V=at => t=V/a
t = 12/480 = 0.025ms
d=0.5at^2
d = 0.5*480*0.000625 =
0.15m = 59"
You missed a decimal place in your conversion from meters to inches.
The problem that I initially found was that 19lb (84N) over a distance of 6" (15cm) only supplies 1.26J. I was surprised I didn't notice that it was exactly 1/10 what it should have been.