Mr. Todd Reeves takes all of the co-op students to Kings Dominion, where they ri

Question

Mr. Todd Reeves takes all of the co-op students to Kings Dominion, where they ride the Volcano.
Assume the 29 kN roller coaster starts from rest and is launched with electromagnetic motors along a horizontal section of track.
It takes 10 m to reach a speed of 35 m/s.

a)What force is applied to the roller coaster during the launch? (N)

b)A meter measures the energy input into the motors during the launch as 0.572 kW-hour. What is the efficiency of the motors? (%)

c)After the launch, the roller coaster goes straight up, shooting out of the mountain. A minimum speed of 7.6 m/s is required at the top. Assuming no losses, how high above the initial elevation can the coaster go? (m)

d)As the coaster comes down the hill to the same elevation at which it started, it loses 29000 J of energy. What is the speed at the bottom of the hill? (m/s)

e)After the run, the roller coaster comes into the station with a speed of 0.44 times the speed at the end of the launch. What percent of energy was lost during the run?

Explanation / Answer

(a)Acceleration of roller coaster= v^2/2s = 35^2/2*10 = 61.25 m/s^2 Time taken to reach that speed = v/a = 35/61.25 = .5714 s. Force applied on roller coaster = mass*acceleration = (29/9.8)*61.25 = 181.25 kN. (b) Energy supplied by motor = .572 kWhr = .5728*3600 kJ = 2059.2 kJ Energy consumed by roller coaster = force*displacement = 181.25*10 = 1812.5 kJ. Efficiency = (1812.5/2059.2 )*100 = 88.02%. (c) Let elevation = h so, 29kN*h = (1/2)*(29/9.8)*(35^2-7.6^2) hence h= 59.55 m. (d) Initial energy = 1812.5 kJ, energy loss = 29 kJ, final energy = 1783.5 kj so velocity v = sqrt(2*1783.5*9.8/29) = 34.71 m/s (e) energy is proportional to v^2 so, Ef = .44^2*Ei = .1936*Ei so percentage energy lost = (1-.1936)*100 = 80.64%

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