You have been hired to design a spring-launched roller coaster that will carry t

Question

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-high hill, then descends 15 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 1.7 m and that a loaded car will have a maximum mass of 400 kg. For safety reasons, the spring constant should be 10% larger than the minimum needed for the car to just make it over the top.

(a) What spring constant should you specify?

(b) What is the maximum speed of a 350 kg car if the spring is compressed the full amount?

Explanation / Answer

Energy from the spring = Energy absorbed by roller coaster to be able to make it to the top of a 11-meter hill

Energy from spring = (1/2)(k)x^2

where

k = spring constant

x = maximum length where spring can be compressed = 1.7 m (given)

For the roller coaster to make it to the top of the 11-m hill, its energy must be

Potential energy = mgh

where

m = 400 kg (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

h = 11 m (given)

Therefore,

(1/2)(k)(1.7)^2 = (400)(9.8)(11)

k = 400*(9.8)*(11)/((1/2)*(1.7)^2)

k = 29840.83 N/m

Since it is required that the spring constant must be 10% larger than the minimum required, then

k = 1.1*29840.83 = 32824.91 N/m

Again, applying the law of conservation of energy,

(1/2)(32824.91)*(1.7)^2 = (1/2)*(350)(V^2)

where

V = maximum velocity of 350-kg car

V = sqrt[(32824.91)*(1.7)^2/((350))] = 16.46 m/sec

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