An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 29 m lower vertically, is a horizontally situated spring with constant 4.1 × 105 N/m.

Question

An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 29 m lower vertically, is a horizontally situated spring with constant 4.1 × 105 N/m.
The acceleration of gravity is 9.8 m/s2 . Ignore friction.
How much is the spring compressed in stop- ping the ore car?
Answer in units of m.

Henry · Answer

V^2 = Vo^2 + 2g*h = 0 + 19.6*29 = 568.4, V = 23.8 m/s.

Work = KE2-KE1 = KE2-0 = KE2.

Work = KE2 = 0.5M*V^2 = 0.5*41,000*(23.8)^2 = 11.7*10^6 J.

Work = F*d = F*29 = 11.7*10^6,
F = 4.03*10^5 N.

d = 1m/4.1*10^5N. * 4.03*10^5N. = = 0.983 m.