A 200-kg roller coaster originally at rest launched off a 12-m high frictionless hill by a spring that is compressed 4 m. If the spring stiffness is 5000 N/m, then how fast is the car going at the bottom of the hill?

Answer 1

energy in spring = initial K.E. of coaster

1/2 k x^2 = 1/2 * 5000 * 4^2 = 40 kJ

K.E. at bottom = initial K.E. + gravitational energy

1/2 * 200 * v^2 = 40 kJ + (m g h) = 40 kJ + (200 * 9.8 * 12)

solve for v