# Work done on two cart system with spring

1. Nov 18, 2015

1. The problem statement, all variables and given/known data
Two identical 0.50-kg carts, each 0.10 m long, are at rest on a low-friction track and are connected by a spring that is initially at its relaxed length of 0.50 m and is of negligible inertia. You give the cart on the left a push to the right (that is, toward the other cart), exerting a constant 4.5-N force. You stop pushing at the instant when the cart has moved 0.50 m . At this instant, the relative velocity of the two carts is zero and the spring is compressed to a length of 0.30 m. A locking mechanism keeps the spring compressed, and the two carts continue moving to the right.

What is the work done by you on the two-cart system?
How far does the system's center of mass move while you are pushing the left cart?
By what amount do you change the system's kinetic energy?

2. Relevant equations

W= force * displacement over which force acts

3. The attempt at a solution
w= 4.5 * 0.5 = 2.25J
I feel like that is not correct because it is only the work done on the one cart, but how do I go about the entire system?

2. Nov 18, 2015

### BvU

Hello Elena,

You may think it's wrong, but it isn't ! The work you do over this 0.5 m is nicely distributed over the two carts and the sping: you haven't just accelerated the cart you were pushing, but (through the spring) also the other one. And you have compressed the spring.

So on to the next step! What relationships do you have to solve that ?

3. Nov 18, 2015

centre of mass moves 0.4m. I got that through reasoning. Since the initial COM is at 25cm, and the final is at 15cm for the spring and the cart moved 20cm to compress the spring to 30 cm, then COM moved 10cm. Then the system continued moving for another 30cm, so the net COM displacement must be 40cm. I was wondering if there was a different approach to this. Something more "physic-sy" for lack of a better word.

And for change in kinetic energy: initial velocity is 0, so Ko=0, and Kf=1/2(m total)(vf)^2. =0.5*(0.5+0.5)(2.121)^2 =2.25J
so ΔK=2.25-0= 2.25J

finding Vf:
Fnet=4.5N=mtotal*a, so a=4.5/(0.5+0.5)=4.5m/s^2.
can use Vf^2=Vo^2+2aΔx
so Vf= √2(4.5)(0.5), Vf≅2.121m/s

Is my thought process for ΔK correct?

4. Nov 18, 2015

I feel like ΔK should not be the same at work.

5. Nov 19, 2015

### BvU

How much energy is there in the spring after the push ?

6. Nov 19, 2015

W=Espring + Ek. Ek=1/2(mtotal)(vf)^2 and Espring=1/2(k)(x^2)

To calculate spring energy I need the k value, right? because Espring=1/2*k*x^2. Since I do not have a k value given in the question, how else could I calculate the energy in the spring?

7. Nov 19, 2015

Or ΔK=Kcm since relative velocity of the two carts is zero.

8. Nov 19, 2015

### BvU

Yes, so I don't think you can come much further than $\Delta E_k = 2.25 {\rm \ J \ }- {1\over 2}k(0.2)^2$

Unless you are supposed to work this all out by integrating the equations of motion ?

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9. Nov 19, 2015

### haruspex

I didn't follow all that, but I don't think that is the right COM displacement. The first cart moves 50cm and the spring compresses by 30cm, so how far is the second cart displaced? What average displacement does that yield?
You mean, 30cm after you stop pushing? That's not how I read the question. Nothing that happens after stopping pushing seems relevant.
I usually avoid using non-inertial frames, but it helps here. Consider the two masses and spring as a subsystem. What is its acceleration relative to the ground?
Using the non-inertial frame of the CoM, what are the forces on each cart?
What is the average compression of the spring?

If you cannot see how to do it that way, just write out the net force versus acceleration equations for each cart and solve.