According to Hooke's Law, $F=-kx$ where $F$ is the restoring force, $k$ is the spring constant and $x$ is the length of extension/compression.
When an applied force compresses a spring, a restoring force will act in the opposite direction.
When a spring is compressed and is in equilibrium (not extending/compressing), the restoring force should be equal to the applied force since the resultant force is 0 and the acceleration is 0. (Thus the spring is not extending/compressing)
However, the act of compressing a spring will require the applied force to be greater than the restoring force even at an infinitesimally small time interval, am I wrong?
Is it more correct to say that $F=kx$ where $F$ is the applied force or $F=-kx$ where $F$ is the restoring force? I think that $F=-kx$ the better option, but I'm pretty confused here. Does it matter at all?
Best Answer
The equation $F=-kx$ is for the spring (restoring) force. If we apply a force equal to this ($F_{app}=kx)$ then the spring edge where the force is applied will not accelerate (it will be stationary or moving at a constant velocity depending on what the velocity is when we set our force to start being equal to $kx$). Note that this is a very specific case
You are right, in order to stretch/compress a spring at rest we need to initially apply a force different than $kx$, but this is true for other scenarios too. For example, to lift a box we need to first apply a force larger than its weight to start it's velocity upwards. Then we can reduce our force to be equal to the weight to lift the box at a constant speed, or we can keep our force constant and the box will accelerate upwards. The same analogy can be made for the spring at each position $x$, since the force depends on $x$.
So overall, the spring (restoring) force is always given by $F=-kx$ where $x$ is the displacement from equilibrium. The applied force is really anything you want it to be, and depending on what it is the spring will respond accordingly based on the initial conditions. For example, we can hang a mass from a spring so that the mass remains stationary, or we can stretch and release that same mass so that oscillations occur. Notice how in either scenario the applied force is a constant $mg$, but the spring behavior is different due to the initial conditions of the system