Question

A single mass m1 = 4.1 kg hangs from a spring in a motionless elevator. The spring is extended x = 13 cm from its unstretched length.

Now, three masses m1 = 4.1 kg, m2 = 12.3 kg and m3 = 8.2 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above.

Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 4.2 m/s2. (Note: an upward acceleration when the elevator is moving down means the elevator is slowing down.)

What is the distance the MIDDLE spring is extended from its unstretched length?

Answer 1

We know that by Hooke's Law,
F = kx; where F is the net force on the spring, k is the spring constant and x is the extension.
We are told that all the springs have the same spring constant as the first, so we first calculate its spring constant.
F = ma = 4.1 × 9.81
= 40.2 Newtons
k = 40.2 ÷ 0.13
k = 309 Newtons / m
Now, for the spring under consideration, the mass is
m2 = 12.3 kg
The net force will be the difference of the downward force of the mass's weight and the upward force of the elevator. Thus,
F = 12.3 × 9.81 - 12.3 × 4.2
F = 69 Newtons
x = 69 ÷ 309
x = 0.22 m = 22 cm