How Does Braking Force Affect Amusement Park Ride Dynamics?

[tucky]

Hi Everyone,

Thank you for dduardo and HllsofIvy for helping me on my last question. Once again, I am having problems with my physics homework. Here is my question:

In an amusement park ride, people in a car are dropped some horrible distance, screaming for their lives. The car rounds a curved track and is brought to a halt be a braking force on a horizontal piece of track. The car a=gas a mass if 1100kg. Point A is 50m off the ground. Point B is 20m off the ground. And point C is 5m off the ground. The track is frictionless until you get to the horizontal braking section. The braking force is 2500lb. 1.) Find the velocity of the car at B and when it reaches the horizontal track. 2.)How far will the car move on the horizontal track? 3.)Then re-work the problem with the braking force kicking in at point B rather than at the horizontal track.

Here is my work, which is more than likely wrong because I have no ideal how what a breaking force is or how to calculate that into my problem:

W = m * g * y
W = 1100kg * 9.8 N * 30 m = 323,400J

W = F * d
323,400J/30m= 10780kg (m/s^2)

a = F/m = 10780kg m/s^2 /1100kg =9.8 m/s^2

v^2=v0^2 +2 average acceleration (x – x0 )
v^2 =0 + 2 (9.8 m/s^2) (30m-0m)
square root of v^2 = square root of (508 m/s)^2
1.) Therefore the velocity at B= 22.5 m/s

W = mv^2 /2 = 1100kg(22.5m/s)^2 /2 = 278,437.5J

2.) W = m * a * d = 278,437.5J * 9.8m/s^2 * d = 25.83m was the distance

3.) I have no ideal how to work the third part.

 

[HallsofIvy]

Looks like a good start: you calculate the potential energy of the car at the start. But why find the force and acceleration? You aren't asked that. You are asked the speed at each of several points. You can find that by "conservation of energy": find the potential energy at points B and C (is that the beginning of the "horizontal" part? You don't say.) Since energy is kinetic energy + potential energy, you must have (1/2)mv^2+ potential energy= potential energy at A or (1/2)m v^2= potential energy at A- potential energy at the new point (B or C). That allows you to calculate the velocity at each point.

Now look at what happens on the horizontal part. There is some "braking" force F= 2500 lb (lb?? POUNDS?? but everything else is in metric. Are you sure it wasn't 2500 N?) and the work done by that is 2500x where x is the distance on the horizontal part. Again, find the potential energy at the end (the kinetic energy is 0 since the car is stopped) and add: 2500x+ potential energy= total energy (which is the same as the same as before).

 

[M98Ranger]

Hi there,

It looks like you are on the right track with your calculations so far. The main concept you need to understand in this problem is the conservation of energy. In this case, the potential energy at point A (50m) is converted into kinetic energy as the car drops to point B (20m). This kinetic energy then remains constant as the car travels along the horizontal track until it reaches point C (5m), where it is converted back into potential energy.

To answer your first question, you correctly used the equation for conservation of energy (W = mgh) to find the initial potential energy at point A. Then, you used the equation for kinetic energy (W = mv^2/2) to find the kinetic energy at point B. However, you made a small mistake in your calculation for velocity at point B. The correct equation to use is v^2 = v0^2 + 2ad, where v0 is the initial velocity (in this case, 0 m/s) and a is the acceleration due to gravity (-9.8 m/s^2). Plugging in the values, we get v^2 = 0 + 2 (-9.8 m/s^2)(50m-20m) = 784 m/s^2. Taking the square root, we get a velocity of 28 m/s at point B.

For your second question, you correctly used the equation for work (W = mad) to find the distance the car moves on the horizontal track. However, you used the wrong value for the acceleration. Since the car is moving at a constant velocity on the horizontal track, the acceleration is 0 m/s^2. Therefore, the distance the car moves is simply d = W/F = 278,437.5J/2500lb = 50.3 m.

For your third question, you need to recalculate the distance using the same equation as before, but with the braking force kicking in at point B instead of the horizontal track. This means that the distance the car travels on the horizontal track will be shorter since the braking force is applied earlier. The work done by the braking force can be calculated as W = Fd, where F is the braking force (2500lb) and d is the distance the car travels before the braking force kicks in. To solve for d, we can use the equation W = mad again, but this time the acceleration is not