# University of Wisconsin Green Bay

An 850 kg roller-coaster is released from rest at Point A of the track shown in the figure. Assume there is no friction or air resistance between Points A and C. How fast is the roller-coaster moving at Point B? What average force is required to bring the roller-coaster to a stop at Point D if the brakes are applied at Point C?

• In this problem, you are first asked to find a speed of a roller-coaster when you are given information about how its height changes.

If non-conservative forces are either known or small and if energy is converted from one form to another between the locations, then any time you relate speed and position of an object at two different points conservation of energy is the most direct way to understand the problem.

In this case, you start out with stored gravitational potential energy and convert part of that energy to kinetic energy.

In the second part of this problem, you are asked to find stopping force given stopping distance. In other words, you want to find how much force is required to convert the roller-coaster’s kinetic energy into heat and sound in a given distance. Again, energy is the most direct way to understand this problem.

• Any time you understand the motion of an object by looking at its energy, you begin with the Conservation of Energy equation. This form of the equation works equally well for the first part of the problem (where Wnc = 0) and for the second question (where you need to solve for Wnc.)

• Step 1

At Point B, part of the gravitational potential energy of the roller-coaster has been converted to kinetic energy, and the roller-coaster has a speed of 30 m/s. Continue to Step 2 to solve for the stopping force required to bring the roller coaster to rest at Point D.

Step 2

An average force of 4200 N is required to stop the roller-coaster by Point D when the brakes are applied at Point C. No further mathematical solution is required for this problem.

• In this problem, you are first asked to find how fast a roller-coaster moves at a given point on the track. The energy conversion chain for this motion is

gravitational potential energy→kinetic energy + gravitational potential energy

Because essentially no energy is lost from the system (you are told friction and air resistance are negligible,) the amount of energy in the system remains constant throughout. This means energy can be compared at any two points. In this case, we are given information about the energy at A and are asked for information B. In other words,

gravitational potential energyA = kinetic energyB + gravitational potential energyB or mghA= ½ mvB2 + mghB as seen in the equation above.
The value of 30 m/s is reasonable for motion of a roller-coaster.

KEA + PEA = KED + PED - Wnc
0 + mghA = 0 + mghD - Wnc
(850 kg)(9.8 m/s2)(140 m) = (850 kg)(9.8 m/s2)(80. m) – F(120 m)cos(1800)
F = 4200 N

In the second part of the problem, you are asked to find the force required to stop the roller-coaster between Points C and D. The energy chain for this motion is

gravitational potential energy→gravitational potential energy + heat and sound or mghA = mghB –Fd cos(1800) as seen in the equation above.
The value of 4200 N is about half the weight of the roller-coaster, so would require a coefficient of friction of about 0.5. Again, a reasonable value.

There are many description of motion problems which can be solved by either kinematics or energy. Energy is almost always the most efficient way to approach these problems as long as enough information is given. (Energy does not require dividing vectors into components, or doing Second Law problems to find acceleration.) In this case, the changing position of the roller-coaster reflected a change in the energy of the system, we are not asked to find time, and there are no significant unknown energy losses from the system. Therefore, we are able to approach this question by tracking the energy.