Three masses $m_1 = 3\text{ kg}$, $m_2 = 9\text{ kg}$ and $m_3 = 6\text{ kg}$ hang from three identical springs in a motionless elevator. The elevator is moving downward with a velocity of $v = -2\text{ m/s}$ but accelerating upward with an acceleration of $a = 5\ \mathrm{m/s^2}$. (Note: an upward acceleration when the elevator is moving down means the elevator is slowing down.) What is the force the bottom spring exerts on the bottom mass? Take $g = 10\ \mathrm{m/s^2}$.
My argument
The elevator is steadily slowing down and it get an upward acceleration.
So as it goes down the net force is
$$\begin{gather}F_s – mg = ma \\
F_s = 6(10 + 5) = 6(15) = 90\text{ N}\end{gather}$$
But this is the force the spring has. It is the same as the force EXERTED on the body attached? Is there a Newton's third law involved?
Friend's argument
He says that BECAUSE it has an upward acceleration and it is going down that we need to be concern with
$$Fs = mg – ma = 90 – 60 = 30\text{ N}$$
Best Answer
If we assume the elevator's motion is adiabatic, so that the springs are never set oscillating, then your answer is correct.
The velocity of the elevator is irrelevant due to the principle of relativity.
The equivalence principle states that when the elevator accelerates, the effects are indistinguishable from those of a gravitational field. Thus, when the elevator accelerates up at $5 m/s$, regardless of its speed, the physics is the same as if the elevator were stationary in a gravitational field whose acceleration is $15 m/s$.
If the elevator's acceleration changes on time scales similar to the damping time of the springs, the masses will oscillate and the force will not be determined from the given information.