How Much Does the Spring Compress When Stopping a 4000 kg Ore Car?

[teggenspiller]

Homework Statement

An ore car of mass 4000 kg rolls downhill on tracks from a mine. At the end of the tracks, at 10 m elevation lower is a spring with k = 400,000 N/m. How much is the spring compressed in stopping the ore car? Ignore friction.

Homework Equations

First of all, I don't think I am visualizing this correctly. why are they telling me its rolling downhill?




F=kx
potential elatic= 1/2kx^2
tmei+Wext=tmef

The Attempt at a Solution

i have the kei=1/2 (4000) *v^2
elastic potential at = 1/2 (400,000) (x) (by the way, we are looking for this 'x'

potential energy= either 0, OR it is mgy 400*9.8*10m


so 2000+200,000 + 39200

= 239400


so all that +wext (work of external forces) = PEf +EPEf+KEf

KEf=0, as said in the question
PEf= the same? seeing as it never went MORE downhill (thats why i am wondering why they ever told me it was originally going down hill

and EPEf= the same! i am still looking for X

so 239400 + 0Wext (?) =

 

[kuruman]

First of all, I don't think I am visualizing this correctly. why are they telling me its rolling downhill?

Because it can't roll uphill (you have to assume it starts from rest, else you cannot do the problem.). But seriously, they want you to see that the initial potential energy when the cart starts out at rest is (finally) converted to elastic energy stored in the spring when the cart is again at rest at maximum compression of the spring.

 

[TobeHode]

potential energy= either 0, OR it is mgy 400*9.8*10m

If the car is 4000kg, this is not the correct potential energy. Find the correct potential energy. Then maybe think about what happens to the potential energy, that is, where does it end up?