Question

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-high hill, then descends 17 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.5 m and that a loaded car will have a maximum mass of 400 kg . For safety reasons, the spring constant should be 14 % larger than the minimum needed for the car to just make it over the top .

A)What is the maximum speed of a 370 kg car if the spring is compressed the full amount?​

Answer 1

Initial elastic potential energy of spring is converted into gravitational energy,

1/2kx^2 = mgy

k = 2mgy / x^2

k = (2 x 400 x 9.81 x 13) / 2.5^2

k = 16323.84 N/m

when spring constant 14% larger, k = 18609.18 N/m

the elaastic potential energy of the spring is final kinetic and potential energy.

1/2kx^2 = mgy + 1/2mv^2

kx^2 = 2mgy + mv^2

when car decents 17 m to the tracks then y = -13 m

v = sqrt[kx^2/m - 2gy]

v = sqrt(18609.18*2.5^2 / 370 - (2*9.81*-13)]

v = 23.86 m/s