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Two blocks of masses $m_1$ and $m_2$ are connected by a mass less spring of constant $k$. The blocks rest on a rough floor with the spring in its equilibrium length. Coefficient of friction between the blocks and the floor is $\mu$. Here we want to find the minimum horizontal force that must be applied on $m_1$ to just move $m_2$. Let $F$ be the force applied to $m_1$ and $X$ the the displacement of the spring when the block $m_2$ starts to move. Hence we have equality $$kX=\mu m_2g.$$ The work done by $F$ is partly dissipated by friction and the rest is stored in the spring as a potential energy. That is

$$FX=\mu m_1gX+\frac{1}{2}kX^2,$$ or equivalently $$F=\mu m_1g+\frac{1}{2}kX.$$ Using the first and the last equation we have $$F=\mu g(m_1+\frac{1}{2}m_2).$$

This solution can be found here. Is this the correct solution? For me this is rather counter intuitive. Let $m_2=10m_1$ then $$F=\frac{3}{5}\mu gm_2.$$ The maximum friction force of $m_2$ alone is $\mu gm_2$, how can a force less than this move the entire system? Furthermore, since $F$ is independent of $k$, we can choose a spring with very large $k$ so that we can view the spring as a rod or very small $k$ so that $m_1$ touches $m_2$. In this case, shouldn't we have $F=\mu g(m_1+m_2)?$ Of course, my intuition could be wrong.

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  • $\begingroup$ For block 1, you are using $\mu$ for the coefficient of kinetic friction and for block 2, you are using $\mu$ for the coefficient of static friction. This aspect of the problem statement doesn't make sense to me. Do they really want you to do that? $\endgroup$ – Chet Miller Feb 29 '16 at 2:44
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    $\begingroup$ For simplification, lets assume the static and kinetic friction coefficient are identical. $\endgroup$ – Sukan Feb 29 '16 at 2:49
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You are right to be suspicious of an answer that contradicts your intuition - but in this case the solution is right

The key (not clearly explained) is that you apply a constant force that is greater than the force needed to move one block: consequently that block will accelerate. As the spring compresses, the first block will accelerate less, and eventually it will decelerate. When it stops, the compression of the spring is maximum - and it applies a greater force to both blocks than the force you used to compress it.

So the solution is correct. If you still have difficulty believing this, think about a hammer: when you swing it, you apply a little bit of force over a long distance; when it hits the nail, it feels a lot of deceleration over a short distance. The force applied to the nail is greater than the force you applied to the hammer. Similar (not same) concept... You do work over a large distance with a little bit of force, to allow you to do the same work over a short distance (with a large force)

Does that help?

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  • $\begingroup$ "... it applies a greater force to both blocks than the force you used to compress it." Doesn't this violate Newton's 3rd Law? If the solution is correct, then for $m_1\ll m_2$ we will have $F\approx\frac{1}{2}\mu gm_2$. That is, using a light object and a spring we can move a heavy object for only half force. I must say this is hard to accept. Anyway, thank you for the explanation. $\endgroup$ – Sukan Feb 29 '16 at 3:26
  • $\begingroup$ No law is violated. If you drop a brick from a tower the force of gravity is mg; when it hits the roof of a car it makes a dent. If you just placed it on the car - no dent. The difference is that the force of gravity acted over a considerable distance and allowed energy to be stored as kinetic energy in the brick. This is what happens here - you store energy in the velocity of the first mass, and release it as the spring becomes more compressed $\endgroup$ – Floris Feb 29 '16 at 3:30
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The point you are missing is that we don't stop expanding or contracting the spring when Kx=F, if we did , then your logic would be correct , here, we are attempting SHM of smaller block here, if we kept the force just less than the required amount, and kept the force acting on m1 , it would shm about mean position and when it's at extreme position, force on m2 would nearly be equal to friction on m2 and when it's a mean postion, since there is extension, force would still be there but less.

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