# Homework Help: Two bars and a spring

1. Jun 9, 2013

### Saitama

1. The problem statement, all variables and given/known data
Two bars of masses $m_1$ and $m_2$ connected by a non-deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to $k$. What minimum constant force has to be applied in the horizontal direction to the bar of mass $m_1$ in order to shift the other bar.

2. Relevant equations

3. The attempt at a solution
The second bar, $m_2$ starts to move when the force due to spring overcomes the force due to friction on $m_2$ i.e $F_{spring}=km_2g$. When $m_2$ starts moving, $m_1$ is still moving. Applying Newton's second law on $m_1$, $F-km_1g-F_{spring}=m_1a$ where F is the minimum constant force asked in the question. I can substitute $F_{spring}$ but that gives me an equation with two variable. How am I supposed to find F from that?

Any help is appreciated. Thanks!

2. Jun 9, 2013

### Staff: Mentor

Good.

Maybe.

Force analysis will only get you so far. (The spring force and acceleration of $m_1$ are not constant.)

Hint: Consider the work done by that constant force.

3. Jun 9, 2013

### Saitama

Why maybe?

From energy conservation,
$$Fx=\frac{1}{2}cx^2+\frac{1}{2}m_1v^2+km_1gx$$
where $x$ is the displacement of $m_1$ and $c$ is the spring constant.

I don't know v but if I put v=0, I do get the right answer but why is v=0?

4. Jun 9, 2013

### Staff: Mentor

Because you want the minimum force F that will just move $m_2$.

5. Jun 9, 2013

### Saitama

Thanks Doc Al!