Question

What is the passenger's apparent weight at t=1.0s?

Answer 1

The apparent weight of the passenger at $t=1\text{ s}$ will be $\boxed{1035\text{ N}}$.

Explanation:

Given:

The mass of the passenger is $75\text{ kg}$.

Concept:

The velocity-time graph of the passenger in an elevator represents the variation of the velocity of the elevator at different instants.

The slope of the velocity time graph of the elevator represents the acceleration and since the velocity-time graph has uniform increasing slope during first two seconds of its motion.

The acceleration of the elevator in the first two seconds is given by:

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{v_2-v_1}{t_2-t_1}$

Substitute the initial and final velocity and time in above expression.

$\begin{aligned}a&=\dfrac{8-0}{2-0}\\&=4\text{ m/s}^2}\end{aligned}$

The apparent weight of the passenger standing in an accelerating elevator is given by.

$W=m(g+a)$

Here, $m$ is the mass of the passenger, $g$ is the acceleration due to gravity.

The acceleration remains constant from time $0\text{ s to }2\text{ s}$ as the slope is constant.

Substitute the values in equation of weight.

$\begin{aligned}W&=75(9.8+4)\text{ N}\\&=1035\text{ N}\end{aligned}$

Thus, the apparent weight of the passenger at $t=1\text{ s}$ is $\boxed{1035\text{ N}}$.

Learn More:

1.  The amount of kinetic energy an object has brainly.com/question/137098

2.  With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward brainly.com/question/9719731

3. You are riding on a roller coaster that starts from rest brainly.com/question/9719731

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

acceleration, velocity, time, graph, passenger, elevator, accelerating, apparent weight, slope, constant, 1035 N, 75 kg.

Answer 2

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight =  m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

Apparent Weight = 1035 N

For downward motion:

Apparent Weight =  m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

Apparent Weight = 435 N