What Is the Passenger's Apparent Weight in an Elevator?

[alexas]

Homework Statement

The figure shows the velocity graph of a 75kg passenger in an elevator.

What is the passenger's apparent weight at t= 1 s ?
What is the passenger's apparent weight at t= 5 s ?
What is the passenger's apparent weight at t= 9 s ?

Homework Equations

F = ma

The Attempt at a Solution

I was thinking since the acceleartion at 1s is 4m/s
The answer for 1s is 300N?

5s, since acceleration is 0, it would be: 0?



Anyways, my answers ended up not being accepted. So i know I am definitely doing something wrong. Any ideas?

 

[PhanthomJay]

We don't see your velocity-teme graph; you first have to determine the correct acceleration, not velocity, at each time interval. Then when you apply Newton's laws, you must first look at all forces acting on the person before using F_net = ma. The 'apparent weight' is the Normal force acting on the person.

 

[alexas]

Heres the picture

 

[alexas]

I was basing the acceleration on the rate of change that was occurring.

 

[LowlyPion]

I was basing the acceleration on the rate of change that was occurring.

It's accelerating because of the increase in speed, this is true, but what about gravity?

 

[myxomatosii]

First let me just say, when you are standing in an elevator, before it begins to move, do you weigh zero?

Think about that for a second, I'm not sure why you would have submitted that but I admit I've had my share of braindead moments myself so I can't be pointing any fingers. Hehehe.

For apparent weight I usually add the acceleration that the system is undergoing TO gravity itself which you must remember (and apparently did not =p ) never goes away.

So yea, think about it.