a roller coster car of mass 1500 kg starts at a distance of H=23m above the bottom of the loop 15 m in diameter. if friction is negligible, what is the downward force of the rails on the can when it is upside down at the top of the loop?(adsbygoogle = window.adsbygoogle || []).push({});

so what i was thinking is this:

f=ma and in this case a=v^2/r so i have .....

f=mv^2/r and i know what m and r are so i need to find v

ei=ef

mgh=mgh+.5mv^2

1500(9.8)23=1500(9.8)15+.5(1500)v^2

v^2=156.8

then i put this into f=mv^2/r and f=3.14 x10^4

is that right?

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# Homework Help: Force of the track on a rollercoaster

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