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.
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?
2 answers
You second solution makes no sense. Why would you set the KE of an object equal to its PE?
1 answer