Solving Freight Car's Initial Speed: Hooke's Law & Friction

[alvarez91]

A 4000kg freight car rolls along rails with negligible friction. The car is brought to rest by a combination of two coiled springs as illustrated in the figure below. Both springs are described by Hooke's law with k1 = 1700N/m and k2 = 3500 N/m. After the first spring compresses a distance of 28.8 cm, the second spring acts with the first to increase the force as additional compression occurs as shown in the graph. The car comes to rest 54.0 cm after first contacting the two-spring system. Find the car's initial speed.

I got the correct answer, .4237, but it took me hours to get, and now I don't even understand how I got it. I understand all the numbers I used, but I just can't figure out which formulas were used. I basically got it by luck. Here is what I did:

.5(4000)v^2=.5(1700).288^2 + (1700).288(.252) + .5(5200)(.252)

I understand the left side and the first part of the right side, but I don't get why I added (1700).288(.252) + .5(5200)(.252). (I understand what all the numbers are, just not why I used them in that order; especially the 5200). Thank you.

 

[PhanthomJay]

The Final energy on the right side of the equation is all spring potential energy, which is the sum of the PE of the first spring (which moves 54 cm) and the PE of the second spring (which moves 25.2 cm).

Your equation will work (if you don't forget to correct your typo and square the the last (.252) term) if you set a = .288, b = .252, and note the final PE is
0.5(1700 (a + b)^2) + 0.5(3500 (b)^2)
0.5(1700 (a^2 + 2ab + b^2) + 0.5(3500 (b)^2)
etc.

If that's what you did.