What spring constant is required to stop the car in a distance of 1.0m?

Question

A 2000-kg car rolls 80m down a frictionless 20 degree incline. If there is a horizontal spring at the end of the incline, what spring constant is required to stop the car in a distance of 1.0m?

Answer 1

First, find the kinetic energy of the car at the bottom of the incline, just before it hits the spring.

The acceleration down the ramp is

a = g*sin(20)
so the displacement is
x = g/2*sin(20)*t^2. solving for the rolling time given that the length of the track is 80m,
t = sqrt(2*80/g/sin(20))

then the velocity at the bottom is

v = g*sin(20)*t = g*sin(20)*sqrt(2*80/g/sin(20))
v = sqrt(2*80*g*sin(20))

the kinetic energy is then given by

K =1/2*m*v^2 = 1/2*m*2*80*g*sin(20) = m*g*80*sin(20)

To stop the car, the spring must absorb all this kinetic energy so

K = U = 1/2*k*x_stop^2, if x_stop = 1 then

k = 2*K = 2*m*g*80*sin(20) = 1.0733e6 N/m

Answer 2

The car has potential energy m*g*h at the start of the roll. You need to do some trig to find h. By the time the car reaches the spring it will have converted all that potential energy to kinetic energy. The spring must do an equal amount of work (Joules) to stop the car where
work = (1/2)*k*x^2

Where does (1/2)*k*x^2 come from? If the spring is compressed a distance x1, the force is k*x1. The average force while the spring is compressed any distance x, is
Fave = (1/2)*k*x
Work is Fave*x