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 le
ngth. 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?
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
The shuttles acceleration in the creases as the fuel is burned because the acceleration of the obect as produced by net force is directly proportional to the magnitude of the net force.
