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?
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.
You second solution makes no sense. Why would you set the KE of an object equal to its PE?
Based on the conversation you provided, it seems that you have two different solutions to calculate the speed of the roller coaster. Let's go through both solutions to check their correctness.
First Solution:
According to bobpursley's suggestion, the potential energy difference between the two hills can be converted into kinetic energy. The equation to calculate kinetic energy is KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
In this case, the potential energy at the top of the second hill is 0.9 PE (assuming the reference point is the first hill). The potential energy at the first hill is given as 18 J. Therefore, the potential energy difference is 0.9 PE - 18 J.
To convert this potential energy difference into kinetic energy, we can equate it to 1/2 * m * v^2:
0.9 PE - 18 J = 1/2 * m * v^2
Since the masses cancel out, you only need to solve for v^2:
0.9 PE - 18 J = 1/2 * m * v^2
0.9 PE - 18 J = 1/2 * 0.035 kg * v^2
Solving this equation will give you the value of v, which represents the speed of the roller coaster.
Second Solution:
According to your suggestion, you can equate the potential energy difference to the gravitational potential energy using the formula mgh. In this case, the potential energy difference is 0.9 PE - 18 J, and the height difference is 25 cm (assuming reference point is the first hill).
Equating mgh to the potential energy difference, we get:
mgh = 0.9 PE - 18 J
0.035 kg * 9.8 m/s^2 * 0.25 m = 0.9 PE - 18 J
Again, solving this equation will give you the value of v, representing the speed of the roller coaster.
To confirm the correctness of your solutions, you can calculate the speed using both methods and compare the results. If they are consistent, you can be more confident in your answer.