Question

You wish to make a simple amusement park ride in which a steel-wheeled roller-coaster car travels down one long slope, where rolling friction is negligible, and later slows to a stop through kinetic friction between the roller coaster's locked wheels sliding along a horizontal plastic (polystyrene) track. Assume the roller-coaster car (filled with passengers) has a mass of 743.0 kg and starts 83.4 m above the ground. (a) Calculate how fast the car is going when it reaches the bottom of the hill. m/s (b) How much does the thermal energy of the system change during the stopping motion of the car

Answer 1

Answer:

(a) The car is going approximately 40.43 m/s at the bottom of the hill

(b) The thermal energy will increase by 607,268.76 J

Explanation:

In the question, we have;

The height of the roller coaster above ground = 83.4 m

The mass of the roller coaster, m = 743.0 kg

(a) By the conservation of energy principle, we have;

The potential energy at the top of the hill, P.E., is equal to the kinetic energy at the bottom of the hill, K.E.

∴ P.E. = K.E.

P.E. = m·g·h

Where;

m = The mass of the roller coaster = 743.0 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height of the roller coaster = 83.4 m

Therefore, we have;

P.E. = 743.0 kg × 9.8 m/s² × 83.4 m = 607,268.76 J

P.E. = 607,268.76 J

K.E. = 1/2·m·v²

∴ K.E. = 1/2 × 743.0 kg × v²

P.E. = K.E.

∴ P.E. = K.E. = 607,268.76 J

1/2 × 743.0 kg × v² = 607,268.76 J

v² = 607,268.76 J/(1/2 × 743.0 kg) = 1,634.64 m²/s²

v = √(1,634.64 m²/s²) ≈ 40.43 m/s

(b) Given that the material wheel moves along polystyrene track, the sound released will be minimal and almost all the kinetic energy will be converted to heat energy when the train stops, therefore, the thermal energy will increase by K.E. = 607,268.76 J

Answer 2

The thermal energy change of the system is 624,492 J.

We know that in the roller coaster, there is an energy transformation from gravitational potential energy to kinetic energy. As such we can write;

mgh = 1/2mv^2

Where we cancel out the mass from both sides, we are left with;

gh=0.5v^2

v= √gh/0.5

v = √10 × 83.4 m/0.5

v = 41 ms-1

Now the kinetic energy is converted also into heat energy hence;

Thermal energy change of the system = 1/2 mv^2 = 0.5 × 743.0 kg × ( 41 ms-1)^2 = 624,492 J

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