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LenaWriter [7]
3 years ago
8

A 2320 pound roller coaster starts from rest and is launched such that it creates a 110 ft high hill with a speed of 65 mph. The

roller coaster travels 1283 ft in reaching the top of the hill and there is a constant drag force of 87 pounds. Determine the required launch energy.
Physics
1 answer:
Sergeu [11.5K]3 years ago
3 0

Answer: E = 941738.537J

Explanation:

to begin,

given that the mass = 2320 pound = 1052.334 kg

Δh = 110 ft = 33.528 m

given that  distance (d) = 1283 ft = 391.058 m

also the speed (v) is 65 mph = 29.058 m/s

force (F) = 87 pounds = 386.995 N

from our knowledge in work energy theory;

E = Fd + 1/2mv² + mgh

E = (386.995 × 391.058) + (1/2×1052.334×29.058²) + (1052.334×9.81×33.528)

E = 151337.491 + 444278.2 + 346122.84

E = 941738.537J

i hope this helps, cheers.

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The formula we use here is:

radial acceleration = ω^2 * R <span>

110,000 * 9.81 m/s^2 = ω^2 * 0.073 m 
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f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm 
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⇒g*t= Vi

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