Solving Frictionless Freight Car's Initial Speed

[laurensummer]

Homework Statement

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 = 1500 N/m and k2 = 3700 N/m. After the first spring compresses a distance of 32.1 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 48.0 cm after first contacting the two-spring system. Find the car's initial speed

Homework Equations

What I did is coming up wrong, but I tried using F = -Kx for both springs and got for F(1) = 1500 * .48 = 720 and for f(2) got 588.8 making the total work 1308.3. I plugged this value into W = 1/2 MV^2 with v = 0.501 in the end. This is not right. I would be so grateful to anyone who could direct me on this one I have been working on it for 2 hours! Its making me crazy!

 

[Stephen Tashi]

To compute work done by a force F(x) that varies with distance, you must compute
$$\int_{d_0}^{d_1} F(x) dx$$
For hooke's law springs, that would be $$\int_{d_0}^{d_1} kx \ dx$$

Are these springs side-by-side or end-to-end with each other?

 

[HallsofIvy]

Homework Statement

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 = 1500 N/m and k2 = 3700 N/m. After the first spring compresses a distance of 32.1 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 48.0 cm after first contacting the two-spring system. Find the car's initial speed

Homework Equations

What I did is coming up wrong, but I tried using F = -Kx for both springs and got for F(1) = 1500 * .48 = 720 and for f(2) got 588.8 making the total work 1308.3.

What? the total force is 1308.3, not work!
Strictly speaking, what you should do is integrate the force function over the stopping distance but because the force function is linear, you can just multiply the average force by the stopping distance.

I plugged this value into W = 1/2 MV^2 with v = 0.501 in the end. This is not right. I would be so grateful to anyone who could direct me on this one I have been working on it for 2 hours! Its making me crazy!

 

[laurensummer]

Stephen, there is a thinner/longer spring (k1) inside of a wider/shorter spring (k2) attached to a wall on the right.
I am going to try what you both have suggested to do and get back, thanks a million.