Relating Potential Energy and Kinetic Energy with springs and inclines

[abpandanguyen]

Homework Statement

A freight car loaded with ore is sitting on a 5^o incline when its brakes fail. After traveling 150 m on a frictionless track, it reaches level ground where it strikes a massive safety spring with a spring constant k = 18,500 N/m. When its velocity has decreased to zero, an automatic latch will catch the car. The equation that can be solved for the car's kinetic energy just before striking the spring is

a. (1/2)mv²_f + mgh_f - (1/2)kx²_f = (1/2)mv²_i + mgh_i - (1/2)kx²_i.

b. (1/2)mv²_f + mgh_f = (1/2)mv²_i + (1/2)kx²_max.

c. (1/2)mv²_f + mgh_f - (1/2)kx²_max = +mgh_i.

d.(1/2)mv²_f + mgh_f = (1/2)mv²_f + mgh_f - (1/2)kx²_f = (1/2)mv²_i.

e. (1/2)mv²_f + mgh_f = (1/2)mv²_i + mgh_i.

Homework Equations

U = mgh
Potential energy of a spring = (1/2)kx²
KE = (1/2)mv²

The Attempt at a Solution

I'm hoping it is right to assume the answer of interest to the question specifically is (1/2)mv²_f as that would be the car's kinetic energy just before striking the spring? (assuming that v_f here is referring to the velocity of the car right before it hits the spring and nothing else)

I'm not quite understanding where to relate the potential energy of the spring into this. I would assume the max potential energy of the spring from compression would be less than the potential energy from the top of the incline.

Wouldn't (1/2)mv²_i be 0 assuming the velocity here is the initial velocity when the car's brakes fail?

Or is this problem simple with the answer being E since the spring is at the bottom of the incline and it would make sense for
(1/2)mv²_f to be equal to mgh_i.

 

[kuruman]

The spring is a red herring. Whether it is there or not, the car will have the same KE at the bottom of the incline.

 

[abpandanguyen]

So I'm hoping E. is right, because that answer seems pretty straightforward >_<..

 

[kuruman]

Can you interpret if what (e) is saying is correct? What is (e) saying to you?

 

[abpandanguyen]

E is telling me that the kinetic energy of it at the point before it hits the spring is equal to the potential energy from its starting position. The other two terms should be 0 since the final potential energy should be 0 and initial kinetic energy (since it is at rest) should also be at 0.

 

[kuruman]

More generally, it is telling you that kinetic plus potential energy at the top is equal to kinetic plus potential energy at the bottom, just before the mass compresses the spring. That's a statement of conservation of mechanical energy. Since we are interested in the speed just before the spring is compressed and this a correct statement of mechanical energy conservation, it is the correct answer.