Max speed with a compressed spring involved

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 18 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.0 m and that a loaded car will have a maximum mass of 420 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.

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

Answer 1

So you would design the car based on max mass of 420 kg. Assuming the velocity is zero at the top of the first hill; that it just makes it to the top, has zero velocity for a fraction of a second, then begins rolling down from that first hill. Then the total energy at the top of that hill is all gravitational potential energy as defined by:

E_GP = mgh = 420(9.81)(13) = 53562.6 joules

The spring potential energy that sent the car up the hill would then also equal 53562.6 joules (if there are assumed to be no losses due to friction, etc)

E_S = (1/2)kx² = (1/2)k(2²) = 53562.6 or k = 26781.3

Add 10% to get k = 29459.43 N/m

So, using k = 29459.43 N/m, we can calculate the spring potential energy:

E_S = (1/2)kx² = (1/2)(29459.43)(2²) = 58918.86 joules.

At the top of the hill, the gravitational potential energy is mgh = (350)(9.81)(13) = 44635.5 joules so the energy still in the car as kinetic energy is 58918.86 - 44635.5 = 14283.36

Since the top of this hill is 18 m below the lowest part of the track, there is an additional gravitational potential energy of (350)(9.81)(18) that the 350 kg car will get from this hill:

(350)(9.81(18) = 61803 joules.

Total energy then at the lowest spot on the track is 61803 + 14283.36 = 76086.36 joules

This is all kinetic energy so:

E_K = (1/2)mv²

(1/2)(350)v² = 76086.36

v² = 434.7792

v = 20.9 m/s or, rounded to 2 sig figs v = 21 m/s