Posted by **Joey** on Thursday, November 4, 2010 at 9:18pm.

This is the question I posted about two hours ago and I worked it out two ways-could you scroll down and check it, please?If I have a roller coaster, and hill 1 height is 50cm and hill height 2 is 25 cm and the PE is .9 as it approaches the top of hill 2, how do I calculate the speed. PE first hill was 18

Please help

Physics-Please help - bobpursley, Thursday, November 4, 2010 at 5:52pm

THe PE is .9 where? Where did the coaster start?

Physics-Please help - Joey, Thursday, November 4, 2010 at 5:54pm

The coaster started at .18PE-the .9 is the PE at the top of the second hill.The second hill is lower.

Physics-Please help - bobpursley, Thursday, November 4, 2010 at 6:01pm

Well, the difference in the two Potential energies is now KEnergy, 1/2 mv^2, solve for v.

Physics-Please check - Joey, Thursday, November 4, 2010 at 6:21pm

. The way to calculate this is you take the KE would be .9. You would say .9 = ½ (.035 x v^2) = 2.215. Correct or no?

Physics-Please check - Joey, Thursday, November 4, 2010 at 7:31pm

Should I have done this as 1/2mv^2 = mgh

Cross out like terms and get to v^2 = square root of 2 x g x .25m(h)= 2.125 m/s

Would this be correct?

- Physics-bobpursley please check -
**bobpursley**, Thursday, November 4, 2010 at 9:33pm
First, you have so many math errors it hurts.

If 1/2 mv^2=.9J

then v^2=2*.9/.035

v=7.17m/s I have no idea where you got 2.215m/s

You have told me several things about the initial PE

a) it was a height of .5m (PE= .035*9.8*.5 = .17J ( you indicated in your solution mass was .035kg, I am not certain where that came from)

b) it was 18J

c) it was .9J at a hill half the height.

So I don't know if it is right or not, I do not know the initial PE.

KE on topsecond hill= InitialPE-PEtopsecondhill

That is the principle you use here.

- Physics-bobpursley please check -
**bobpursley**, Thursday, November 4, 2010 at 9:34pm
You second solution makes no sense. Why would you set the KE of an object equal to its PE?

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