# Find the car's initial speed

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1. Feb 18, 2015

### Meagan

1. The problem statement, all variables and given/known data
A 5000-kg 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 and have spring constants with k1 = 1700 N/m and k2 = 3000 N/m. After the first spring compresses a distance of 28.5 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 45.0 cm after first contacting the two-spring system. Find the car's initial speed

2. Relevant equations
I do not know.

3. The attempt at a solution
Completely wrong so please help!

#### Attached Files:

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2. Feb 18, 2015

### Nathanael

Show your work anyway. Did you try using conservation of energy?

3. Feb 18, 2015

### Meagan

I tried using Fs=-kx, then adding all of the forces together to find the work with Ws=∑Fr, and then using the work-kinetic energy theorem, I was not exactly sure how to set this equation up correctly, however, to find the initial velocity.

4. Feb 18, 2015

### Nathanael

It's a valid approach, but the force is not constant. You need to show your work so I can see where you go wrong.
Do it in two pieces; for the first 28 cm there is a certain force acting, and for the rest of the distance there is another force acting.

5. Feb 18, 2015

### Meagan

I was thinking there was technically three forces, the first spring by itself, the second spring, and then the first spring again but over a different distance. I did: F1=-1700 * 0.285m, F2=-3000*0.165m and then F3=-1700*0.165m. I left out the units for k to avoid confusing that anymore than necessary.

6. Feb 18, 2015

### Meagan

Using this approach I found the net work to be 567 J. I am not really sure how to transition from this information to the work-kinetic energy theorem. I tried W=1/2 (k1+k2)vf2-1/2 k1vi2 and solving for initial velocity.

7. Feb 18, 2015

### lightgrav

because the Force is not constant, the Work is ½ Fmax xmax = ½ k s2 ... where s is the spring stretch (or compression)

8. Feb 18, 2015

### Nathanael

You have the right idea. Using 0.165*(k1+k2) is the same as breaking it into two pieces like you did 0.165*k1+0.165*k2

But 1700*0.285 and (3000+1700)*0.165 is only the force at those distances! The force varies with distance... The formula Work = Force * distance only applies to constant forces... The correct formula is work = ∫F.dx

9. Feb 18, 2015

### Meagan

And in this situation

Ok. I will try again after my next class. Thank you for your help and sorry for my initial mess up not showing my work. This was my first post lol.

10. Feb 18, 2015

### Meagan

I do not understand how to use that formula in this situation. I only remember how to integrate when there is a function.

11. Feb 18, 2015

### Nathanael

The function is F=-k1x for the first 28.5 cm and F=-(k1+k2)x for the remaining distance.

In other words, the integral (for the work) is the area under the graph in post #1 from zero to 45 cm.

12. Feb 19, 2015

### Meagan

Well I got it wrong . That's ok though, I think I know what I did wrong! Thank you for all of your help!

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