# Homework Help: Force acting on a spring (rubber band)

1. Dec 15, 2012

### kotchenski

1. The problem statement, all variables and given/known data
A person performs an exercise where he pulls a rubber band stuck in a wall to later return it to its resting position. The two movements forward and backward, is equivalent to one practice cycle. They both happen in a horrizontal plane with a constant velocity of v=0.5ms-1. The hand which grabs the band has the coordinate x=0 when resting and the coordinate x=0.5m when fully stretched. The bands' springconstant is k=480Nm-1

2. Relevant equations
Write (as a function of x) the force F1(x) the person has to perform in order to pull the band towards himself. Furthermore write the force F2(x) needed to return it to its resting position. The equations has to be explained referral to relevant physical laws.

3. The attempt at a solution
Since the movement happens in a horrizontal plane, I can exclude the normal force and the force of gravity since they exclude each other. Fn+Ft=0 → Fn=-Ft
Since the band is being stretched and compressed, it must be proportional to the force excerted on it. Meaning that it must follow Hooke's law. I can thereby write the force F1 as a function of x as:
F1(x)=-kx
When the person pulls the band with the force F1(x) an equal and opposite force is trying to return it to its resting position, this force must be F2(x). Thereby:
F2(x)=-F1(x)
Reffering to relevant physical laws, there are no better examples than newton's laws.
It follows newton's 1st law since it experiences no net force because it moves at constant velocity and because it moves in a straight horrizontal line. From this I can point out newton's 2nd law, it experiences no net force since it moves with constant velocity, and because of that it has no acceleration which means Fnet=ma → ma=0.
Finally there's newton's 3rd law. When the person is pulling the band, he exerts a force F1 while simultaniously a force F2 exerts a force F2=-F1 therefor the system experiences two forces that are equal in magnitude but oppositely directed.
Would this be sufficient as an answer or is there something missing/something wrong?

2. Dec 15, 2012

### Staff: Mentor

I don't like the style of the question. It is an overly complicated problem statement for a simple answer - and at the same time, it leaves some ambiguity how the answer should look like.
As F2 points in the same direction as F1, I think I would use the same sign for both forces. On the other hand, the force needed to return it is 0 - you can just let it go, it will return to its resting position on its own (neglecting the vertical movement).

F=-kx is just Hooke's law.