Janushka S.

asked • 06/21/19

Work, power and energy

Heights:

Point A=65m, Point B=52 m, Point C=25m

The total mass of the roller coaster is 650 kg.

a) If it takes a force of 2760 N [parallel to the track] to pull the roller coaster up to point A, and it is pulled along 150 m of track, find the work done on the roller coaster. 

b) Find the total energy the roller coaster has at point A if it is at rest for a brief moment before it starts to go down the hill. 

c) According to the Law of Conservation of Energy, what will the total energy of the roller coaster be at points B and C? 

d) Use the Law of Conservation of Energy to find the speed of the roller coaster at point B. 

e) If it takes the roller coaster 3 minutes to travel around the track, find the power of the roller coaster. 

f) In real life, some of the energy the roller coaster has at point A will be lost as thermal energy due to friction as it travels through the track. If the speed of the roller coaster at point C is only 19 m/s, calculate the amount of energy that was lost due to friction. 

2 Answers By Expert Tutors

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Stanton D. answered • 06/26/19

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Osmaan S. answered • 06/24/19

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Stanton D.

Agree with values except for part A: the work done on the car is 414 kJ. Gravity did *not* do work on the car. The net work done is reflected in the (gravitational) potential energy gain of the car, after the pull to point A. Also, in part E, it's unclear what the actual net total displacement & path "around the track" (over which the power should be calculated) of the car was. We only know points initial (and no altitude on that!), A, B, and C. Can only hope that the car did not hit a barrier at point C and eject all passengers forward at 19 m/s.
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06/26/19

Osmaan S.

I respectfully disagree, in Part A, gravity certainly did do work on the car. Gravity will always do work on an object as long as there’s a net change in height. In Part A, the car changed height by 65 m, so since height changed, gravity must have done work. This can be easily calculated by using the generalized form of work: W = Fd cosθ where F is the force (gravity, pointing straight down), d is the displacement vector, and θ is the angle between F and d. Here, θ is an angle between 90 and 180 degrees, so cosθ will be negative, and F and d are non-zero, so gravity clearly does work, and the work done by gravity is equal and opposite to the work done by the car. Therefore, the net work was zero. We can also verify this by the work-energy theorem: the net work done on an object is equal to the change in kinetic energy of the object. Assuming the car started at 0 m/s, which is a reasonable assumption, we know that at Point A, it also came to rest (as given in the problem). Therefore, the initial and final velocities are the same (0 m/s) so the change in kinetic energy is zero, and therefore the net work done must also be zero. As for Part E, you’re right in that we don’t know the displacement, but the problem says that the coaster takes 3 minutes to give fully around the track, so the displacement must be zero. We just need the total work done by the car, and the time taken. The time was given, and I assumed that the only work done on the car was the one that I calculated (since they didn’t give us any more information). Honestly, I think this part of the problem could have been more clear in its description.
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06/26/19

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