An ore car of mass 4,700 kg rolls downhill on tracks from a mine.At the end of the tracks,11.1 m lower in elevation,is a spring with k = 410,000 N/m.How much is the spring compressed in stopping the ore car? Ignore friction.

Question

An ore car of mass 4,700 kg rolls downhill on tracks from a mine.At the end of the tracks,11.1 m lower in elevation,is a spring with k = 410,000 N/m.How much is the spring compressed in stopping the ore car? Ignore friction.
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Quizplus · Accepted Answer

The potential energy of the ore car is converted to spring potential energy when the car reaches the spring. So, we can use the conservation of energy principle to solve this problem. 
Initial potential energy = mgh1 
Final potential energy = 0 (at the spring's equilibrium position) and spring potential energy = 1/2 kx^2
Equating both: 
mgh1 = 1/2 kx^2 
x^2 = 2mgh1/k
x = sqrt(2mgh1/k) = sqrt(2*4700*9.81*11.1/410000) = 1.58 m 
Therefore, the spring is compressed by 1.58 meters in stopping the ore car. Hence, the answer is C.

An Ore Car of Mass 4,700 Kg Rolls Downhill on Tracks