Question

A single mass m₁ = 3.5 kg hangs from a spring in a motionless elevator. The spring is extended x = 15 cm from its unstretched length. Spring constant is 228.9

2) What is the distance the lower spring is stretched from its equilibrium length?

3) Now the elevator is moving downward with a velocity of v = -2.7 m/s but accelerating upward with an acceleration of a = 5.5 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?

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

5) Finally, the elevator is moving downward with a velocity of v = -3 m/s and also accelerating downward at an acceleration of a = -1.6 m/s².

is the elevator speeding up? slowing down? moving at a constant speed?

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

x₁ = x₂ = x₃

x₁ > x₂ > x₃

x₁ < x₂ < x₃

₇₎ What is the distance the MIDDLE spring is extended from its unstretched length?

Answer 1

2). What is the distance the lower spring is stretched from its equilibrium length?

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 * 15 = 30cm

3)

4.Now the elevator is moving downward with a velocity of v = -2.7 m/s but accelerating upward with an acceleration of a = 5.5 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?

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) do calculation for final answer

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

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)

u have not given the values m2 and m3. so i'm just giving the formulas.

Again, using Hooke's law:

x = F/ 3.3e2

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

c) x1 < x2 < x3

7)The distance the MIDDLE spring is extended from its unstretched length when not accelerated is 45cm