To determine the force exerted by the scale on a 53-kg person standing in various situations within an elevator, we need to consider the effective acceleration acting on the person in each case. The force exerted by the scale (which is the normal force) can be found using Newton's Second Law of Motion, [tex]\( F = ma \)[/tex].
Given:
- The mass of the person, [tex]\( m = 53 \, \text{kg} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
Let's analyze each scenario:
### a. The elevator moves up at a constant speed.
When the elevator is moving at a constant speed, there is no change in velocity (no acceleration apart from gravity). Thus, the only force exerted by the scale is the gravitational force (weight):
[tex]\[ F_{\text{up, constant speed}} = m \cdot g \][/tex]
### b. It slows at [tex]\( 2.0 \, \text{m/s}^2 \)[/tex] while moving upward.
In this case, the elevator is decelerating while moving upward. The net acceleration is the difference between gravitational acceleration and the deceleration of the elevator:
[tex]\[ a_{\text{up, slowing}} = g - 2.0 \, \text{m/s}^2 \][/tex]
The force exerted by the scale will be:
[tex]\[ F_{\text{up, slowing}} = m \cdot (g - 2.0) \][/tex]
### c. It speeds up at [tex]\( 2.0 \, \text{m/s}^2 \)[/tex] while moving downward.
While moving downward and accelerating, the effective acceleration is the sum of gravitational acceleration and the acceleration of the elevator:
[tex]\[ a_{\text{down, speeding}} = g + 2.0 \, \text{m/s}^2 \][/tex]
The force exerted by the scale will be:
[tex]\[ F_{\text{down, speeding}} = m \cdot (g + 2.0) \][/tex]
### d. The elevator moves down at a constant speed.
When the elevator moves downward at a constant speed, there is no change in velocity (no net acceleration apart from gravity). Thus, the force exerted by the scale is simply the weight of the person:
[tex]\[ F_{\text{down, constant speed}} = m \cdot g \][/tex]
### e. It slows to a stop while moving downward with a constant acceleration of [tex]\( 2.5 \, \text{m/s}^2 \)[/tex].
Here, the elevator is decelerating while moving downward. The effective acceleration is the gravitational acceleration minus the deceleration of the elevator:
[tex]\[ a_{\text{down, slowing}} = g - 2.5 \, \text{m/s}^2 \][/tex]
The force exerted by the scale will be:
[tex]\[ F_{\text{down, slowing}} = m \cdot (g - 2.5) \][/tex]
Using the given values, the forces calculated for each scenario are as follows:
1. [tex]\( F_{\text{up, constant speed}} = 53 \cdot 9.8 = 519.4 \, \text{N} \)[/tex]
2. [tex]\( F_{\text{up, slowing}} = 53 \cdot (9.8 - 2.0) = 413.4 \, \text{N} \)[/tex]
3. [tex]\( F_{\text{down, speeding}} = 53 \cdot (9.8 + 2.0) = 625.4 \, \text{N} \)[/tex]
4. [tex]\( F_{\text{down, constant speed}} = 53 \cdot 9.8 = 519.4 \, \text{N} \)[/tex]
5. [tex]\( F_{\text{down, slowing}} = 53 \cdot (9.8 - 2.5) = 386.9 \, \text{N} \)[/tex]
Therefore, the forces exerted by the scale on the person during each situation are:
a. 519.4 N
b. 413.4 N
c. 625.4 N
d. 519.4 N
e. 386.9 N
