Calculating Spring Compression of Ore Car on Downhill Mine Tracks

[Idividebyzero]

1.An ore car of mass 38000 kg starts from rest
and rolls downhill on tracks from a mine. At
the end of the tracks, 6.5 m lower vertically,
is a horizontally situated spring with constant
5.2 × 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?



2. W=F*D W=1/2 kx²



3. Started with 1/2kx₂ the constant is given to us. work is also F*D here once it leaves the hill I considered it in free fall. so using force of gravity as F.
(m)(g)(d)=1/2kx² and solved for x. this was incorrect. thought about the work-energy principle but i don't know any kinematic inforation other than g and the vertical displacement. any thoughts

 

[SammyS]

The method looks good.

What do your calculations look like? What did you get for an answer?

 

[clementc]

Yeah there doesn't seem to be anything wrong with it =\

 

[Idividebyzero]

here is what i got

W=1/2kx^2

mgd=1/2kx^2

(38000kg)(9.8m/s^2)(6.5m)=1/2(5.2e5)x^2
(2420600 kgm^2/s^2)=260000*x^2
9.31 = x^2
3.05 = x

using the correct sig figs would round to 3.1 and the answer was wrong

 

[Idividebyzero]

weird, just tried submitting 3.05 m and it was correct. that was the un-rounded raw answer before sig figs. My professor made it clear to use sig figs when doing the assignment. :anger: