# Trouble With Springs-1

1 vote
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Question-
Two blocks of masses m1 and m2 are connected by a spring of force constant K. Then, the minimum force F to be applied to m1 in order to just slide m2 is
(coefficient of friction is μ)

My Attempt-
I consider the block m1, block m2 and the spring to be the system on which the force is applied. Then, one has to apply a minimum force only to overcome the static friction between the blocks and the surface. The spring force developed becomes an internal force.
F = μm1g + μm2g

F = μg(m1 + m2)

But the right answer is slightly different from this. What mistake have I commited?

I think your reasoning is correct. What is the "right" answer? Is there any explanation?
The Right Answer is ${\rm{F = }}\mu g\left( {{m_1} + {\rm{ }}\frac{{{m_2}}}{2}} \right)$
The same question was posted on Physics Stack Exchange : [Blocks spring system](http://physics.stackexchange.com/q/240528).
The person who asked the question isn't seemed to be satisfied with the Floris's answer. I'm too not satisfied with it. I just want to know what is wrong with the above reasoning?

1 vote

The same question was posted on Physics Stack Exchange : Blocks spring system. It contains a good explanation.

The key point is that the force which causes $m_2$ to move is not $F$, it is $F_2=\mu m_2 g$ just as you expect. $F_2$ equals $F$ plus the rate of change of momentum of $m_1$.

You are thinking of the situation in which the two blocks are initially in contact. In that case $m_1$ cannot move until $m_2$ also moves, and the force needed to move both is $\mu (m_1+m_2)g$. In the present case $m_1$ has accelerated over some distance, building up momentum. It then collides with $m_2$ via the spring. The force between the two blocks during the collision equals $F$ plus the rate of change of momentum of $m_1$.

Another analogy is to think of a policeman trying to force open a locked door, which requires a force $F_2$ to break the lock. If he leans against the door and pushes against it the maximum force he can apply is the static friction force $F_1$ between his shoes and the ground. However, if he takes a run up to the door and collides with it he can apply more force. The force which accelerates the policeman is still the friction force $F_1$, but when this is applied over a distance it builds up and stores momentum $P$ in the policeman. If the policeman bounces off the door elastically in time $\Delta t$ the force actually applied to the door is then $F+\frac{2P}{\Delta t}$. The 2nd term is the rate of change of momentum of the policeman, and can be much larger than $F$.

answered Mar 23, 2017 by (26,660 points)
In the link provided by you - http://physics.stackexchange.com/questions/240528/blocks-spring-system
Shouldn't conservation energy equation also include the $\text{Kinetic Energy}$ of the first block?
Time as a Function of Compression?