https://file.aiquickdraw.com/mathbot/file/172849748831916e9ocai.jpg

To calculate the weight of the 68 kg passenger in the elevator at \( t = 1 \, \text{s} \), we need to consider the forces acting on the passenger at that time.

### Step 1: Recognize the situation
The weight of a person in an elevator depends on the apparent weight, which can change if the elevator is accelerating. The apparent weight is the normal force \( F_N \) exerted by the floor, which can differ from the gravitational weight \( W \). 

- The gravitational weight \( W \) is given by:
  \[
  W = mg
  \]
  where \( m = 68 \, \text{kg} \) is the mass and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

### Step 2: Determine velocity and acceleration at \( t = 1 \, \text{s} \)
From the velocity-time graph, we can see that at \( t = 1 \, \text{s} \), the elevator is still accelerating since the velocity is increasing. The slope of the velocity graph gives us the acceleration.

Between \( t = 0 \) and \( t = 2 \, \text{s} \), the velocity increases from 0 to approximately \( 8 \, \text{m/s} \). So the acceleration \( a \) is:
\[
a = \frac{\Delta v}{\Delta t} = \frac{8 \, \text{m/s} - 0}{2 \, \text{s}} = 4 \, \text{m/s}^2
\]

### Step 3: Calculate the net force
The net force \( F_{\text{net}} \) acting on the passenger is:
\[
F_{\text{net}} = ma = 68 \, \text{kg} \times 4 \, \text{m/s}^2 = 272 \, \text{N}
\]

### Step 4: Calculate the apparent weight
The apparent weight (normal force) \( F_N \) is the sum of the gravitational force and the net force due to acceleration. When the elevator is accelerating upwards, the apparent weight is:
\[
F_N = W + F_{\text{net}} = mg + ma = 68 \, \text{kg} \times 9.8 \, \text{m/s}^2 + 272 \, \text{N}
\]
\[
F_N = 666.4 \, \text{N} + 272 \, \text{N} = 938.4 \, \text{N}
\]

Thus, the weight of the passenger at \( t = 1 \, \text{s} \), or the apparent weight, is approximately **938.4 N**.

Would you like more details or have any other questions?

Here are 5 related questions to expand the topic:
1. How does the apparent weight change when the elevator decelerates?
2. What would the apparent weight be at \( t = 6 \, \text{s} \) when the elevator is moving at constant velocity?
3. How would the apparent weight change if the elevator was accelerating downwards instead?
4. What would the passenger's apparent weight be if the elevator were in free fall?
5. How can we calculate the work done by the elevator on the passenger?

**Tip:** The slope of a velocity-time graph gives the acceleration, which plays a key role in determining forces in motion scenarios.

What is the weight of a 68 kg passenger in an elevator at t = 1s?

Determine the apparent weight of a 68 kg passenger in an elevator at t=1s, considering the forces involved while the elevator accelerates upwards. Using Newton's Second Law, we calculate the net force and normal force, resulting in an apparent weight of approximately 938.4 N.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation