Question

A single mass m1 = 3.6 kg hangs from a spring in a motionless elevator. The spring is extended x = 15.0 cm from its unstretched length. . . Now, three masses m1 = 3.6 kg, m2 = 10.8 kg and m3 = 7.2 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 distance the lower spring is stretched from its equilibrium length?

Answer 1

To determine the distance the lower spring is stretched from its equilibrium length, use Hooke's Law and the weight of the mass to calculate the force exerted by the spring. Then, divide this force by the spring constant to find the distance the spring is stretched.

Step-by-step explanation:

To determine the distance the lower spring is stretched from its equilibrium length, we can use Hooke's Law which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed from its equilibrium position. Since the spring constants are the same for all three masses, the distance the lower spring is stretched can be calculated by dividing the force exerted by the spring (calculated using the weight of the mass) by the spring constant.

Let's calculate:

1. Mass m1 = 3.6 kg, Spring constant k = ?
   Using Hooke's Law: F = kx, where F is the force exerted by the spring and x is the distance the spring is stretched. Rearranging the equation, we have k = F/x. We can find F by multiplying the mass m1 by the acceleration due to gravity (9.8 m/s^2). Using the given displacement x = 15.0 cm = 0.15 m, the force constant of the spring can be calculated.
2. Once we have calculated the spring constant k, we can use it to calculate the distance the lower spring is stretched for masses m2 and m3. We can use the same formula F = kx and solve for x by rearranging the equation as x = F/k.

By following these steps, we can determine the distance the lower spring is stretched from its equilibrium length for the given masses.

Answer 2

The distance the lower spring is stretched from its equilibrium length is 45cm because the weight is 3x as much as the reference spring and the spring constant is the same.

2) The force the bottom spring exerts on the mass is its weight (=mg) PLUS 10.8kg x 3.8m/s^2 = 133N

3) The distance the upper spring is extended from its unstretched length when not accelerated is 15cm

4) Rank the distances the springs are extended from their unstretched lengths:
c) x1 < x2 < x3

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

6) Finally, the elevator is moving downward with a velocity of v = -3.4 m/s and also accelerating downward at an acceleration of a = -2.1 m/s2.
a)speeding up
because the v and a are in the same direction