# Roller Coaster Problem

1. Nov 28, 2004

### intrigue

Hello, everyone. I'm new here at the forums, and I've got a little physics problem I'd like your input on:

"A roller coaster reaches the top of the steepest hill with a speed of 6.0 km/h. It then descends the hill, which is at an average angle of 45 degrees and is 45 m long. What will its speed be when it reaches the bottom? (Assume the coefficient of kinetic friction = 0.12)"

I believe this is intended to be an inclined plane problem, but I think that either the mass or the weight of the roller cart is necessary to determine the rate at which it accelerates down the hill. Is there some way of solving the problem with the information given or is it impossible to do so without somehow determining the mass of the cart? Thanks in advance.

-Intrigue-

PS It was hinted that we ought to convert km/h to m/s before attempting to solve the problem, but this does nothing to suggest the mass or the weight of the roller coaster. On an ideal coaster (without friction) it might be possible to calculate the weight of the cart based on its speed at the top of the hill with respect to the side it climbs before the descent (assuming that both sides are 45 meters long and at the same 45 degree angle) but the coefficient at the end of the problem assures the existence of at least some friction, so this idea doesn't work either. Please post your thoughts.

2. Nov 28, 2004

### Parth Dave

Try it out. Mass will cancel out when you do it. Draw yourself a free body diagram and find the net forces in the horizontal and vertical direction. Then see if you can find your horizontal acceleration.

3. Nov 28, 2004

### intrigue

Of course! F=ma, so a=Fnet/m. Fx=mgcos45, and since the force of friction is equal to the coefficient of friction times the normal force (which is equal to Fy and also Fx in this case, since we're dealing with a 45-45-90 right triangle) it ends up equaling mgcos45 (the normal force) times the coefficient of friction, which was .12, as given in the equation.

We're left with another basic F=ma problem, if I did all of this right, except this time we remember that Fnet = Fx - Force of Friction:

a = F/m = [mgcos45 - (.12)mgcos45]/m

All three masses cancel, so we're left with:

gcos45-(.12)gcos45, which equals roughly 6.1 m/s^2

Then, using a constant acceleration problem involving distance, velocity, and acceleration and solving for velocity...

I got 23 m/s for the velocity at the bottom of the hill. Does that sound right to you?

4. Nov 29, 2004

### James R

You can also work this out using conservation of energy, if you like.

5. Nov 29, 2004

### intrigue

Do you get the same result?