At the bottom of one loop a roller coaster carriage (mass 1500 kg) has a kinetic energy of 4.32 x 10^5 J. If the loop has a diameter of 27 m, can it make it to the top of the circular loop ( assuming no friction loss)?
2007-05-21 07:05:26 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Hi. The coaster will slow by 9.8m/s so you must find the speed at the bottom. The rise in height (27m) determines the minimum speed (velocity).
2007-05-21 07:09:41 · answer #1 · answered by Cirric 7 · 0⤊ 0⤋
KE should be greater then or equal to PE to make it to the top of the circular loop.
2007-05-21 07:19:02
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answer #2
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answered by Yssa A 3
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PE = mgh
= 1500 kg (9.8 m/s^2)(27m)
= 3.969 x 10^5 J
PE
ok, KE = PE, and PE(potential energy)= mgh.
So since KE = PE, you can say that 4.32 x 10^5 = mgh, where m=1500kg, and g=9.8 m/s^2 solving for h, you get h= 29.388 m, which is more than the 27 m high that the loop is, so yes, the coaster can make it.
2007-05-21 07:13:09 · answer #3 · answered by DesertFox33 2 · 0⤊ 0⤋
[5/2]R is right... submit to in recommendations that on the suited of the curler coaster, top is 2R, no longer R (as a results of fact the suited is two radii intense). replace this in to your 2nd skill power and you get [5/2]R, that's right.
2016-11-25 22:13:48 · answer #4 · answered by ? 4 · 0⤊ 0⤋