Find the force which makes the box jump

[kaspis245]

Well, I think the upward net force acting on $m_1$ in position $x_1$ is the same force which creates a downward net force in position $x_2$, so maybe this applies:
$kx_1-m_1g+F=kx_2+m_1g$

Then I used these equations:

$F+m_1g=kx_1$
$kx_2=m_2g$

And got $F=\frac{g(m_1+m_2)}{2}$.

Does this happen to be the correct answer?

 

[TSny]

Well, I think the upward net force acting on $m_1$ in position $x_1$ is the same force which creates a downward net force in position $x_2$, so maybe this applies:
$kx_1-m_1g+F=kx_2+m_1g$

The left side should represent the upward net force at $x_1$ after the applied force F is removed.

 

[kaspis245]

Sorry, $kx_1-m_1g-F=kx_2+m_1g$. Now I see that it gets me nowhere...

 

[TSny]

You still don't have the left side correct. There are only two forces that act on $m_1$ after F is removed.

 

[kaspis245]

Sorry again: $kx_1-m_1g=kx_2+m_1g$.

 

[TSny]

OK. This gives you a third equation that you can use along with your two equations from post #25.

 

[kaspis245]

I get the answer $F=g(m_1+m_2)$. Is it correct?

 

[TSny]

I get the answer $F=g(m_1+m_2)$. Is it correct?

Yes. I believe that's the answer. The only thing lacking is a justification for why you can claim that the net force on $m_1$ at $x_1$ (after F is removed) is equal to the net force at $x_2$.

 

[kaspis245]

That's actually quite hard. Does it have something with spring potential energy?

 

[TSny]

Conservation of energy is another way to relate $x_1$ and $x_2$. That's actually the first way I worked the problem.

But you got me thinking about just the forces. Have you studied simple harmonic motion (SHM)? If so, then you can think of $m_1$ as moving in SHM between the lowest and highest points. It's a property of SHM that the acceleration of the moving mass has the same magnitude at the extreme positions ($x_1$ and $x_2$, in this case). Same acceleration implies same net force.