An 840 Kg roller coaster car is launched from a giant spring of constant k=31 kN/m into a frictionless loop-the-loop track of radius 6.2 m, as shown below. What is the minimum amount that the spring must be compressed if the car is to stay on the track?

Answer 1

At the top of the loop mv^2/R=mg+N

However if we ant minimum speed put N to zero.

Thus to stay in the loop mv^2/R=mg

v at the top thus = sqrt(g/R)

To get to the top of the loop you are at a height of 2R, so the amount of energy lost getting there is 2mgR.

Thus the initial energy must be of magnitude 2mgR + mv^/2 so

kx = 2mgR + mg/2R

Divide by K and you're all set!