# Two blocks with spring (no friction), force applied: displacement?

1. Sep 6, 2011

### ddtozone

1. The problem statement, all variables and given/known data

A 2kg block & 3kg block (left & right, respectively) are on a horizontal frictionless surface connected by a spring with K=140 N/m. A 15N force is applied to the 3kg block towards the right. How much does the spring stretch from its equilibrium length?

2. Relevant equations

F=-kx
others, maybe?

3. The attempt at a solution

I simply did force over constant, which is 15/140=0.107m. Is it this simple, or do you use the masses in some way?

Thank you!

2. Sep 6, 2011

### Rayquesto

the gravitational force of the 3kg is going to be more than the 2 kg gravitational force. So, when you pull with 15N to the right, it will not act like a fixed spring constant. So, what you should do is say 3-2kg=1kg so the resultant will include 9.81 newtons in terms of the spring being moved, not the entire system which includes the spring and the two blocks. So, I think the answer is 15N-9.81N then you multiply by its constant.

3. Sep 6, 2011

### ddtozone

Oh ok, that makes more sense. But wouldn't you divide by the constant instead of multiplying by the constant, so the final answer would be N/(N/m)=m?