A single mass m1 = 3.1 kg hangs from a spring in a motionless elevator. The spri

Question

A single mass m1 = 3.1 kg hangs from a spring in a motionless elevator. The spring is extended x = 12 cm from its unstretched length.

1) What is the spring constant of the spring? 252.425 N/m Submit

2) Now, three masses m1 = 3.1 kg, m2 = 9.3 kg and m3 = 6.2 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? 182.466 N Submit

3) What is the distance the lower spring is stretched from its equilibrium length? 1789.99 cm Submit

4) Now the elevator is moving downward with a velocity of v = -3.6 m/s but accelerating upward with an acceleration of a = 3.9 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? N

5) What is the distance the upper spring is extended from its unstretched length? cm

6) Finally, the elevator is moving downward with a velocity of v = -3.1 m/s and also accelerating downward at an acceleration of a = -3.3 m/s2. The elevator is: speeding up slowing down moving at a constant speed

7) Rank the distances the springs are extended from their unstretched lengths:

x1 = x2 = x3

x1 > x2 > x3

x1 < x2 < x3

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

Explanation / Answer

(1) Fg= kx

3.1(9.8) =K(0.12)

k=253.17 N/m

(2)The top spring is supporting all the three masses,so it is exerting a force equal to force of gravity of all masse

F = (m1 + m2 + m3) * g

= 18.6 * 9.8

F= 182.3 N

(3)The lower spring is supporting only mass m3, so Fg= kX3

also m3=2m1

so X3= 2X1

X3= 2(12)

X3= 24 cm

(4)F3= m (g+a)

F3= 6.2 (9.8+3.9)

F3=84.9 N

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