Homework Help: Physics

Posted by Tammy on Wednesday, November 14, 2007 at 8:02pm.

A roller coaster car of mass 1000kg is riding along a track from point A to point C. (point A is 75m high, B is 50m high and C is 100m high, so that there are a total of three hills.) the final velocity at point C is 10m/s

a) Assume no friction what is the speed of the car at the top of hill A?

intial = final
1/2mv_a^2 +mgh_a = 1/2mv_c^2 + mgh_c
1/2V_a^2 + 750 = 1050
V_a = 24.5 m/s

b) what is the max speed the car can achieve along the track?

What would be the equation for this part of the problem?

c) assume the car starts with the same speed as a) and there is friction present. The car does 50000J of work while the car moves from A to C, will the car still be able to ascend to the top of hill C? At what speed is the car at the top of C?

W= change in KE
and I got V_c = 26.5m/s

d) If the distance along the track from point A to point C is 200m what is the average magnitude of the force of friction between the track and the car?

I don't know how to find this part of the problem can you please help me?

Physics - bobpursley, Wednesday, November 14, 2007 at 8:13pm
cant you use the energy lost on the track (avgfrictionforce*distance) to find it? You know the starting and ending energies...

Intial PE+ initial KE= final PE+ final KE + frictionforce*total distance.

on b), use energy again.
Physics - drwls, Wednesday, November 14, 2007 at 8:21pm
a)You seem to be using g = 10 m/s^2 instead of the actual 9.8 m/s^2. If you are going to carry three significant figures, you should use a g that is that accurate. You also do not seem to have used h_c of 100 m in your g h calculation.

b)To get the highest speed, you need information on the lowest elevation. You have not provided that information.

c) To do this, you need the accurate starting velocity at A, but I am not sure you did part a) correctly to obtain that value

d) Divide the 50,000J work done (which I assume is frictional) by the distance travelled (200 m) to get the average force.

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