An ore car of mass 4 000 kg rolls downhill on tracks from a mine.At the end of the tracks,10.0 m lower in elevation,is a spring with k = 400 000 N/m.How much is the spring compressed in stopping the ore car? Ignore friction.
A)0.14 m
B)0.56 m
C)1.40 m
D)1.96 m

Question

Quizplus · Accepted Answer

The potential energy of the ore car at the top of the hill is converted to the potential energy of the compressed spring at the bottom of the hill.

Using conservation of energy, we can calculate the initial potential energy of the ore car:

PE_initial = mgh
PE_initial = (4000 kg)(9.81 m/s^2)(h)
PE_initial = 392040 J

where h is the height of the hill.

At the bottom of the hill, all of this potential energy is stored in the compressed spring:

PE_final = (1/2)kx^2

where x is the compression of the spring.

Setting PE_initial equal to PE_final, we can solve for x:

392040 J = (1/2)(400000 N/m)(x^2)
x^2 = 1.96
x = 1.4 m

Therefore, the spring is compressed by 1.4 m in stopping the ore car. The answer is C.

An Ore Car of Mass 4 000 Kg Rolls Downhill