A single mass (m 1 = 3.1 kg) hangs from a spring in a motionless elevator. The s

Question

A single mass (m1 = 3.1 kg) hangs from a spring in a motionless elevator. The spring constant is k = 344 N/m.

1)What is the distance the spring is stretched from its unstretched length?

Now, three masses (m1 = 3.1 kg, m2 = 9.3 kg and m3 = 6.2) hang from three identical springs in a motionless elevator. The springs all have the same spring constant given above.

What is the magnitude of the force the bottom spring exerts on the lower mass?

3)What is the distance the middle spring is stretched from its equilibrium length?

4)Now the elevator is moving downward with a velocity of v = -3.1 m/s but accelerating upward at 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 magnitude of the force the upper spring exerts on the upper mass?

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

6)Finally, the elevator is moving downward with a velocity of v = -3.5 m/s but accelerating downward at an acceleration of a = -2.2 m/s2.

Compare the magnitude of the NET force on each mass:

F1 = F2 = F3

F1 > F2 > F3

F2 > F3 > F1

7)What is the magnitude of the net force on the middle mass?

Explanation / Answer

Since there is a top spring, middle and bottom, then it must be that:

s1 is attached to the ceiling (elevator) and m1 hangs from it;
s2 is attached to m1; m2 hangs from s2;
s3 is attached to m2; m3 hangs from s3.

1)

F = k * x is Hooke's Law. Applied to this situation, where the force is gravity:

Fg = m * g = k * x

k = m1 * g / x

x = m1*g / k

x = 3.1 x 9.8 / 344

x = 0.088 m

x = 8.8 cm

2)

Fg = (m1 + m2 + m3) * g

Fg = 18.6 * 9.81

Fg = 182.47 N

3)

The lower spring s3 is supporting only m3; applying Hooke's law:

x = Fg / k = m * g / k

Notice, however that m3 = 2 * m1, therefore:

x3 = 2 * x1 = 2 * 8.8 = 17.6 cm

4)

The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s3 is supporting its mass m3 to balance gravity.

As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.

When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:

F3net = m3 * (g + a)

= 6.2 * (9.8 + 3.9)

= 84.94N

5)

Assuming this question applies to the decelerating elevator, the upper spring s1 is not only balancing the force of gravity on all 3 masses but also counteracting the decelerating force:

F = (m1 + m2 + m3) * (g + a)

F = 18.6 x 13.7

F = 254.8 N

Again, using Hooke's law:

x = 254.8 / 344

x = 0.740 m = 74 cm

6)

Assuming again that this question applies to the decelerating elevator, the middle spring s2 is not only balancing the force of gravity on m2 and m3 but also counteracting the decelerating force:

F = (m2 + m3) * (g + a)

F = (9.3 + 6.2) * (9.8 + 3.9)

x = F / k

x = (9.3 + 6.2) * (9.8 - 3.9) / 344

x = 0.2658m

x = 26.58 cm

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