A single mass m 1 = 4.6 kg hangs from a spring in a motionless elevator. The spr

Question

A single mass m1 = 4.6 kg hangs from a spring in a motionless elevator. The spring is extended x = 10 cm from its unstretched length. What is the spring constant of the spring? --> 450.8 N/m. Now, three masses m1 = 4.6 kg, m2 = 13.8 kg and m3 = 9.2 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 force the top spring exerts on the top mass? --> 270.48 N. What is the distance the lower spring is stretched from its equilibrium length? --> 20 cm. Now the elevator is moving downward with a velocity of v = -2.4 m/s but accelerating upward with an acceleration of a = 4.3 m/s2. (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? -->129.72 N. What is the distance the upper spring is extended from its unstretched length? --> 86.32 cm.

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

Explanation / Answer

considering all three masses as a single system.

Fnet = ma

F1 - (m1 + m2 + m3)g = (m1 +m2 +m3)a

F1 = (m1 + m2 + m3)(a+g)

F1 = (4.6 + 13.8 + 9.2) ( 4.3 + 9.81) = 389.44 N

now using Fnet = ma on lower mass,

F3 - m3g = m3a

F3 = 9.2 ( 9.81 + 4.3) = 129.81 N

for upper spring, F1 = kx

389.44 = 450.8x

x = 0.8639 m = 86.39 cm

now for middle spring, using Fnet = ma
now using Fnet = ma on middle mass,

F2 - F3 - m2g = m2a

F2 - 129.81 - (13.8 x 9.81) = 13.8 x 4.3

F2 = 324.53 N

or it can be found using on upper mass,

F1 - F2 - m1g = m1a

389.44 - F2 - 4.6*9.81 = 4.6*4.3

F2 = 324.53 N

using F2 = kx

324.53 = 450.8x

x = 0.72 m = 72 cm

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