Question

replace the rope with an ideal spring of the same length that holds the same block.

1. which of the following is true about the force of the spring when the elevator is gradually slowing down?

a. $f_{spring} > mg$
b. $f_{spring} < mg$
c. $f_{spring} = mg$
d. $f_{spring} = mg$

1. what will happen to the length of the spring as the elevator gradually comes to a stop?

a. the length of the spring increases so new length $> x_1$
b. the length of the spring decreases so new length $< x_1$
c. the length of the spring stays the same.
d. the length of the spring will be 0.

Explanation:

Question 3

When the elevator is gradually slowing down (assuming it's moving upward, decelerating, or moving downward, accelerating upward), the acceleration of the block (and elevator) is upward. Using Newton's second law \( F_{net}=ma \), where \( F_{net}=F_{spring}-mg \) (taking upward as positive). Since \( a>0 \) (upward acceleration), \( F_{spring}-mg = ma>0 \), so \( F_{spring}>mg \). Wait, but if the elevator is slowing down while moving downward, the acceleration is upward too. Wait, maybe the initial state: when elevator is at rest, \( F_{spring}=mg \). When slowing down (decelerating), if moving upward, acceleration is downward? Wait, no: acceleration direction is opposite to velocity when slowing down. If elevator is moving upward and slowing down, acceleration is downward. If moving downward and slowing down, acceleration is upward. Wait, the problem says "gradually slowing down" – maybe we assume the elevator is moving upward (common scenario). Wait, no, let's re-examine. Wait, the options: if \( F_{spring}>mg \), net force upward (acceleration upward); \( F_{spring}

Case 1: Elevator moving upward, slowing down: velocity \( v \) upward, acceleration \( a \) downward (since slowing down). So net force on block: \( mg - F_{spring}=ma \) (downward net force). Thus, \( F_{spring}=mg - ma < mg \).

Case 2: Elevator moving downward, slowing down: velocity \( v \) downward, acceleration \( a \) upward (slowing down). Net force: \( F_{spring}-mg = ma \) (upward net force), so \( F_{spring}=mg + ma > mg \).

But the problem says "gradually slowing down" – maybe the default is moving upward? Wait, the original problem (before this) – maybe the elevator was moving upward? Wait, the question is about replacing a rope with a spring, holding the same block. When elevator is slowing down (decelerating), if moving upward, acceleration is downward, so spring force is less than mg. If moving downward, acceleration upward, spring force more than mg. But the options include \( F_{spring}mg \) (option a). Wait, maybe the elevator is moving upward and slowing down, so acceleration downward, so \( F_{spring}=mg - ma < mg \), so option b is correct? Wait, but maybe I messed up. Wait, let's use Newton's second law: \( F_{net}=ma \). The forces on the block are spring force upward (\( F_{spring} \)) and weight downward (\( mg \)). So \( F_{net}=F_{spring}-mg \) (upward positive). If acceleration \( a \) is downward (elevator moving upward, slowing down), then \( a \) is negative (if we take upward as positive). So \( F_{net}=F_{spring}-mg = m(-a) \) (since \( a \) downward is \( -a \) upward). Thus, \( F_{spring}=mg - ma < mg \). So \( F_{spring}

Wait, but maybe the elevator is moving downward and slowing down: acceleration upward, so \( F_{net}=F_{spring}-mg = ma \) (a upward), so \( F_{spring}=mg + ma > mg \), option a. But the problem doesn't specify direction, but in typical problems, when slowing down, if not specified, maybe moving upward. But the option b is circled, maybe that's the intended answer. Wait, the problem's option b is \( F_{spring}

The spring force is \( F_{spring}=kx \) (Hooke's law, \( x \) is extension from natural length). From question 3, if \( F_{spring}

Answer:

b. \( F_{\text{spring}} < mg \)

Question 4