Compressing a spring So i just learned that the force exerted from a spring equals the spring constant times the length of the spring:
$$ F_s=k~\ell $$
However, this would mean that if you were to compress a spring with your hands, you would feel the greatest amount of resistance in the beginning, because as soon as the length of the spring decreases, so would the force it exerts, and it would just accelerate inwards until it breaks. This is obviously not what actually happens, as a real spring would simply reach equilibrium and exert an equal force to the one applied (provided the force is not too great, of course) after being compressed a little. So my question is why does the force of the spring increase when the length decreases, when the formula says the force should decrease?
 A: The equation for the spring is subtly different from the one you give. It is actually:
$$ F_s=k~\Delta\ell $$
where $\Delta\ell$ is the change in the spring length.
Suppose you take a spring a metre long and compress it by a millimetre ($10^{-3}$m) then the force is not $k$ times one metre, it is $k$ times the change in the length i.e. $k$ times a millimetre in this case:
$$ F_s=k \times 10^{-3} $$
As you compress the spring more and more the change in length gets bigger and bigger, so the force gets greater and greater. Which is of course exactly what we observe.
A: The length you mentioned in the formula is actually the displacement from the equilibrium position and not simply the length of the spring.
The actual formula is $F = -k\Delta l$.
This is the restoring force of the spring on account of the inertia of the spring.
 So, as the spring is stretched, $\Delta l$ is positive and the restoring force is negative. The case is opposite for compression of the spring.
Thus, exactly what accounts is the displacement of the spring from its equilibrium position that determines the amount of restoring force.
If you look at the magnitude of the force, then the force is proportional to the change in the length of the spring from the equilibrium position.
