**Physics Problem: Elevator Deceleration and Apparent Weight****Problem Description:**You are inside an elevator that is descending with decreasing speed, i.e., the elevator is decelerating. Compared to your actual weight, your apparent weight is:(a) greater. (b) lesser. (c) same. (d) could be greater or lesser. (e) None of the above is correct. **Analysis:**When an elevator is descending and decelerating, it means that although the elevator is still moving downward, its speed is reducing. This situation results in an upward acceleration of the elevator. To analyze the apparent weight in this scenario, we need to consider the forces acting on a person inside the elevator.### Forces Acting on You:1. **Gravitational Force (Weight):** This force acts downward and is equal to \( mg \), where \( m \) is your mass and \( g \) is the acceleration due to gravity.2. **Normal Force (Apparent Weight):** This is the force exerted by the floor of the elevator on you and acts upward.### Newton's Second Law in Action:- When the elevator is in free fall or moving with constant speed (without deceleration or acceleration), your apparent weight equals your actual weight.- However, when the elevator decelerates while descending, there is an upward net acceleration. According to Newton’s second law, \( F = ma \).### Calculation of Apparent Weight:- The net force acting on you in the upward direction is given by \( N - mg = -ma \), where \( a \) is the upward acceleration (or negative downward acceleration).- Rearranging the equation, the normal force (apparent weight) \( N \) becomes \( N = mg + ma \). - Since \( a \) (the elevator's deceleration) is positive in this context, your apparent weight becomes \( mg + ma \), which is greater than \( mg \).### Conclusion:When the elevator is decelerating (moving downward but with decreasing speed), your apparent weight becomes **greater** than your actual weight.Thus, the correct answer is:**(a) greater.**

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Description: **Physics Problem: Elevator Deceleration and Apparent Weight****Problem Description:**You are inside an elevator that is descending with decreasing speed, i.e., the elevator is decelerating. Compared to your actual weight, your apparent weight is:(a

(a) greater The apparent weight is if elevator going downwards and slowing down: N = mg + m|a|.…

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. You are inside an elevator that is descending with decreasing speed, i.e., the elevator is decelerating. Compared to your actual weight, your apparent weight is (a) greater. (b) lesser. (c) same. (d) could be greater or lesser. (e) None of the above is correct.

. You are inside an elevator that is descending with decreasing speed, i.e., the elevator is decelerating. Compared to your actual weight, your apparent weight is (a) greater. (b) lesser. (c) same. (d) could be greater or lesser. (e) None of the above is correct.

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**Physics Problem: Elevator Deceleration and Apparent Weight**

**Problem Description:**
You are inside an elevator that is descending with decreasing speed, i.e., the elevator is decelerating. Compared to your actual weight, your apparent weight is:

(a) greater.
(b) lesser.
(c) same.
(d) could be greater or lesser.
(e) None of the above is correct.

**Analysis:**
When an elevator is descending and decelerating, it means that although the elevator is still moving downward, its speed is reducing. This situation results in an upward acceleration of the elevator. To analyze the apparent weight in this scenario, we need to consider the forces acting on a person inside the elevator.

### Forces Acting on You:
1. **Gravitational Force (Weight):** This force acts downward and is equal to \( mg \), where \( m \) is your mass and \( g \) is the acceleration due to gravity.
2. **Normal Force (Apparent Weight):** This is the force exerted by the floor of the elevator on you and acts upward.

### Newton's Second Law in Action:
- When the elevator is in free fall or moving with constant speed (without deceleration or acceleration), your apparent weight equals your actual weight.
- However, when the elevator decelerates while descending, there is an upward net acceleration. According to Newton’s second law, \( F = ma \).

### Calculation of Apparent Weight:
- The net force acting on you in the upward direction is given by \( N - mg = -ma \), where \( a \) is the upward acceleration (or negative downward acceleration).
- Rearranging the equation, the normal force (apparent weight) \( N \) becomes \( N = mg + ma \).
- Since \( a \) (the elevator's deceleration) is positive in this context, your apparent weight becomes \( mg + ma \), which is greater than \( mg \).

### Conclusion:
When the elevator is decelerating (moving downward but with decreasing speed), your apparent weight becomes **greater** than your actual weight.

Thus, the correct answer is:
**(a) greater.**

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