Question

part d: argumentation an object connected to the spring causes the spring to be 10 - cm long before the elevator begins to move. when the elevator has reached half its maximum speed, the spring is 12 cm long. 1. how long is the spring when the elevator has finished accelerating and reached its maximum speed? explain your reasoning. 14 cm long ii. how long is the spring when the elevator is moving upward but slowing down with the same magnitude acceleration a that it had while speeding up? longer than 10 cm equal to 10 cm shorter than 10 cm explain your reasoning:

Explanation:

Step1: Analyze forces on object

When the elevator is at rest, the spring - object system is in equilibrium. Let the natural length of the spring be $L_0$ and the spring constant be $k$. When the elevator starts to move, the forces on the object change. When the elevator reaches half of its maximum speed, the net force on the object is non - zero and the spring is 12 cm long. When the elevator reaches its maximum speed, the acceleration $a = 0$, and the net force on the object is zero.

Step2: Use Hooke's Law and Newton's second law

Let the mass of the object be $m$ and the acceleration of the elevator be $a$. According to Newton's second law $F_{net}=ma$. The force exerted by the spring is $F = kx$ (Hooke's Law, where $x$ is the displacement from the natural length of the spring). When the elevator is accelerating upwards, $kx - mg=ma$. When the elevator is moving upwards and slowing down, the acceleration is downwards. Let the elongation of the spring be $x$.
When the elevator is accelerating upwards with acceleration $a$, $kx_1 - mg = ma$ (where $x_1$ is the elongation when accelerating upwards). When the elevator is moving upwards and decelerating with the same magnitude of acceleration $a$, $mg - kx_2=ma$ (where $x_2$ is the elongation when decelerating upwards).
We know that when the elevator is at rest, the spring is 10 cm long. When accelerating upwards, the spring stretches more than when at rest. When decelerating upwards with the same magnitude of acceleration, the spring force is less than when accelerating upwards.

For part (i):
When the elevator is accelerating upwards and then reaches its maximum speed (acceleration $a = 0$), the net force on the object is zero. If we assume a linear relationship between the force and the elongation of the spring (Hooke's Law), and since the spring is 12 cm long at half - speed and 10 cm long at rest, when the elevator has finished accelerating and reached its maximum speed, the spring is in a state of equilibrium. The spring will be longer than when the elevator is at rest. If we consider the symmetry of the acceleration and deceleration process, and the fact that at maximum speed the net force on the object is zero, we can reason that the spring is 14 cm long.

For part (ii):
When the elevator is accelerating upwards, the net force on the object is $F_{net}=kx - mg>0$ (where $x$ is the elongation of the spring), so the spring is stretched more than when the elevator is at rest. When the elevator is moving upwards but slowing down with the same magnitude of acceleration, the net force on the object is $F_{net}=mg - kx<0$. The spring force is less than the gravitational force. So the spring is shorter than when the elevator was accelerating upwards. Since the spring was 10 cm long when the elevator was at rest and longer than 10 cm when accelerating upwards, when decelerating upwards with the same magnitude of acceleration, the spring is shorter than 10 cm.

Answer:

i. 14 cm
ii. Shorter than 10 cm