Calculating Area Under a Curve with Multiple Springs

[Bashyboy]

I attached the graph, provided in the problem, as a document.

This is the wording of the problem: 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 = 3500 N/m. After the first spring compresses a distance of 28.2 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 55.0 cm after first contacting the two-spring system. Find the car's initial speed.

I am trying to write a general formula for find the area under this curve. This is what I have come up with: \frac{1}{2}k_1x^2_1 + k_1x_1(x_2-x_1) + \frac{1}{2}[k_1x_2+k_2(x_2-x_1)](x_2-x_1) But this just doesn't seem correct. The graph is rather confusing.

 

[SteamKing]

Your area calculation looks good. You can simplify it some if you look up how to calculate the area of a trapezoid. (the part of the curve from x1 to x2)

 

[Basic_Physics]

Your formula is correct.
The force is determined via Hooke's law: F = kx
What confuses you most likely is the 1 and 2 subscripts for the x distances. x1 is just the compression distance where the second spring starts to compress and x2 is just another compression point beyond x1, so at x2 both springs are compressed, but spring 2 is compressed only by x2 - x1 , while spring one is compressed by the "total" distance x2.

 

[Bashyboy]

I got 413 J as the area under the curve, does that seem right? I have another question: is the work done by the springs on the train the same amount of work done by the train on the springs? I wonder, because the area I am calculating under the curve is the net work done by the springs system on the train; and would this be equal to the initial kinetic energy of the train, before in comes in contact with the springs?

 

[haruspex]

\frac{1}{2}k_1x^2_1 + k_1x_1(x_2-x_1) + \frac{1}{2}[k_1x_2+k_2(x_2-x_1)](x_2-x_1)

Your second term is wrong by a constant factor. A simpler method is to calculate the total energy stored in each spring separately in the final state.

 

[Bashyboy]

How exactly would I do that?

 

[Bashyboy]

If I had found the area of the trapezoid, rather than the sum of the areas of the rectangle and right-triangle, would the formula for the area look like this:

\frac{1}{2}k_1x_1^2+\frac{(k_1x_2)+(k_1x_2+k_2(x_2-x_1))}{2}(x_2-x_1)

 

[Bashyboy]

Okay, I calculated the work done by the springs on the car:

\frac{1}{2}(0.282)(1700)(0.282)+\frac{[(1700)(0.282)+(3500)(0.268)]}{2}(0.268)

The springs did 257.527 J of work on the car. I figured that this meant that the change in kinetic energy of the car was 0 J - 257.527 J (the final kinetic energy being 0 J and initial being 257.527 J). This led me to \Delta K = -1/2mv_i^2. After the required algebraic manipulation, to find the initial velocity, and substitution of the values, the equation yielded 0.321 m/s. The actual value is 0.391. What did I do wrong?

 

[haruspex]

If I had found the area of the trapezoid, rather than the sum of the areas of the rectangle and right-triangle, would the formula for the area look like this:

\frac{1}{2}k_1x_1^2+\frac{(k_1x_2)+(k_1x_2+k_2(x_2-x_1))}{2}(x_2-x_1)

Still doesn't look right to me.
When things come to rest, how much is spring 1 compressed? So what is the energy in it? Similarly spring 2. (I don't think you should end up with any terms involving both k1 and x1.)

 

[Basic_Physics]

Sorry, after looking at the formula in your first post again I realized it is wrong. The perpendicular height of the last triangle is the difference between the two forces so all that is left for the last term is
\frac{1}{2}(k_{1}+k_{2})(x_{2}-x_{1})^{2}
I get the total work done 383 J then (the work done will be negative actually since the restoring force and displacement are in opposite directions).

 

[haruspex]

The perpendicular height of the last triangle is the difference between the two forces so all that is left for the last term is
\frac{1}{2}(k_{1}+k_{2})(x_{2}-x_{1})^{2}

Yes, and you can see that in the total formula all terms involving both k1 and x1 cancel.