Below is a roller coaster design. The 500 kg coaster enters the power life with

Question

Below is a roller coaster design. The 500 kg coaster enters the power life with a speed of 5.24 m/s at a height of 22.5 meters. It is then lifted to a height of 124.5m in 25.0 seconds and exits the lift at point 2 with a speed of 8.26 m/s. The coaster rolls down the incline and enters a loop. After the loop, the coaster is brought safely to rest by a large spring that compresses 30.0 meters. Find (a) power needed to lift the coaster from point 1 to 2 (b) the speed of the coaster when it enters the loop (c) the acceleration of the coaster as it enters the loop (in g's)

Explanation / Answer

work done in lifting the coaster from point 1 to 2 = mgh

work done = 500 * 9.8 * (124.5 - 22.5)

work done = 499800 J

power = work / time

power = 499800 / 25

power = 19992 W

by conservation of energy

initial energy = final energy

mgh + 0.5 * mv^2 = 0.5 * mv1^2

gh + 0.5 * v^2 = 0.5 * v1^2

9.8 * 124.5 + 0.5 * 8.26^2 = 0.5 * v1^2

v1 = 50.084 m/s

speed of the coaster when it enters the loop = 50.084 m/s

let the radius of the loop be r so

acceleration = v^2 / r

acceleration = 50.084^2 / (r * 9.8) g

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