I got the answer right, but it involved some guessing. So I’m here to make sure I got a conceptual understanding of this.
Normal force is a contact force. If the car was not in contact with the loop (or barely in contact), the loop would exert no normal force on the car. So at the minimum...
Okay. I treated the system as all 3 blocks accelerating together at 3.8 m/s^2, since the top spring is carrying all three blocks.
Fnet,system = (m, system)(a) = Force of upper spring - Weight of all three blocks.
With these calculations I got a downwards displacement of 1 meter, which is the...
m1 = 4 kg, m2 = 12 kg, m3 = 8 kg. k = 327 N/m for all three blocks. The elevator accelerates upwards at 3.8 m/s^2.
Net force of block one would be equal to force applied by top spring minus weight of system, since top spring is holding all 3 blocks.
F1 = 4*3.8= Fs,top - Wsystem = Fs,top -...