Question

A single mass m1 = 3.6 kg hangs from a spring in a motionless elevator. The spring is extended x = 15.0 cm from its unstretched length. . . Now, three masses m1 = 3.6 kg, m2 = 10.8 kg and m3 = 7.2 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above. . . What is the distance the lower spring is stretched from its equilibrium length?

Answer 1

The distance the lower spring is stretched from its equilibrium length is 45cm because the weight is 3x as much as the reference spring and the spring constant is the same. 

2) The force the bottom spring exerts on the mass is its weight (=mg) PLUS 10.8kg x 3.8m/s^2 = 133N 

3) The distance the upper spring is extended from its unstretched length when not accelerated is 15cm 

4) Rank the distances the springs are extended from their unstretched lengths: 
c) x1 < x2 < x3 

5) The distance the MIDDLE spring is extended from its unstretched length when not accelerated is 45cm 

6) Finally, the elevator is moving downward with a velocity of v = -3.4 m/s and also accelerating downward at an acceleration of a = -2.1 m/s2. 
a)speeding up 
because the v and a are in the same direction