It takes the elevator in a skyscraper 5.0 ss to reach its cruising speed of 14 m/sm/s. A 80 kgkg passenger gets aboard on the ground floor. What is the passenger's weight before the elevator starts moving? What is the passenger's weight while the elevator is speeding up?What is the passenger's weight after the elevator reaches its cruising speed?

To solve this problem, we need to understand how the passenger's weight changes in different situations based on the elevator's motion. The weight experienced by the passenger is the normal force, which varies depending on the acceleration of the elevator. We'll calculate the weight at three different stages: 

1. **Before the elevator starts moving:**
   - The elevator is stationary, so the passenger's weight is simply the gravitational force acting on them.
   - The weight \( W \) is given by:
     \[
     W = mg
     \]
     where:
     - \( m = 80 \, \text{kg} \) is the passenger's mass,
     - \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

     Substituting the values:
     \[
     W = 80 \times 9.8 = 784 \, \text{N}.
     \]
     So, the passenger's weight before the elevator starts moving is **784 N**.

2. **While the elevator is speeding up:**
   - When the elevator is accelerating, the passenger experiences a net force in addition to gravity. The total force on the passenger (normal force, which they perceive as their weight) will increase due to the elevator's acceleration.
   - The elevator takes \( 5.0 \, \text{s} \) to reach its cruising speed of \( 14 \, \text{m/s} \), so the acceleration \( a \) is:
     \[
     a = \frac{\Delta v}{\Delta t} = \frac{14 \, \text{m/s}}{5.0 \, \text{s}} = 2.8 \, \text{m/s}^2.
     \]
   - The net force experienced by the passenger (their perceived weight) is now:
     \[
     W = m(g + a).
     \]
     Substituting the values:
     \[
     W = 80 \times (9.8 + 2.8) = 80 \times 12.6 = 1008 \, \text{N}.
     \]
     So, while the elevator is speeding up, the passenger's weight is **1008 N**.

3. **After the elevator reaches its cruising speed:**
   - Once the elevator reaches its constant cruising speed, the acceleration is zero. Hence, the weight reverts to the gravitational force alone.
   - The weight is the same as before the elevator started moving:
     \[
     W = 784 \, \text{N}.
     \]
     So, the passenger's weight after the elevator reaches cruising speed is **784 N**.

### Summary:
1. **Before the elevator moves**: 784 N.
2. **While the elevator is accelerating**: 1008 N.
3. **After the elevator reaches cruising speed**: 784 N.

Would you like more details on any part of this? 

### Related Questions:
1. How would the passenger's weight change if the elevator were decelerating instead of accelerating?
2. What would happen to the passenger's weight if the elevator were accelerating downward?
3. How would the time to reach cruising speed affect the perceived weight during acceleration?
4. What is the net force on the passenger during the different stages of the elevator's motion?
5. How does Newton's second law apply to the changes in the passenger's weight?

**Tip**: Always remember that the perceived weight changes due to the normal force, which depends on the elevator’s acceleration or deceleration.

It takes the elevator in a skyscraper 5.0 s to reach its cruising speed of 14 m/s. A 80 kg passenger gets aboard on the ground floor. What is the passenger's weight before the elevator starts moving? What is the passenger's weight while the elevator is speeding up? What is the passenger's weight after the elevator reaches its cruising speed?

Learn how to calculate a passenger's weight before, during, and after an elevator's acceleration in a skyscraper. This problem covers Newton's Laws of Motion, forces, and kinematics, providing detailed steps for students in Grades 9-12. Includes key formulas like W = mg and W = m(g + a).

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