Elevator Problem Apparent weight and distance traveled

[charan1]

Homework Statement

Henry gets into an elevator on the 50th floor of a building and it begins moving at t=0s. The figure shows his apparent weight over the next 12 seconds.

Part A: Is the elevator's initial direction up or down? Explain how you can tell.

Part B: What is Henry's mass?

Part C: How far has Henry traveled at t=12s?

Homework Equations

F=ma
F=m(g + ay)

The Attempt at a Solution

Part A:

The elevator is initially going in the downward direction. This is because the apparent weight stays the same for 2 seconds if it went upwards the apparent weight would have gone up right then starting from t=0.

Part B:

First I just did F=ma at 600N because that was the start...so 600N=mg. m came out to be 61.22kg. This was incorrect.


Then I thought that whenever there is an apparent weight change that is when the elevator switches direction because it has to break equilibrium in the forces to switch the net forces direction. So I suppose that since the apparent weight is 750N mostly I put that in force and it was: 750N=mg so m= 76.53kg

Part C:

Im not sure how to do this. I suppose there is some relation between apparent weight change and acceleration? like Fnet=ma and the a will be the acceleration for the direction change?

The only thing is the image does not show how long the acceleration is for.

please help for B and C

 

[pgardn]

Homework Statement

Henry gets into an elevator on the 50th floor of a building and it begins moving at t=0s. The figure shows his apparent weight over the next 12 seconds.

Part A: Is the elevator's initial direction up or down? Explain how you can tell.

Part B: What is Henry's mass?

Part C: How far has Henry traveled at t=12s?

Homework Equations

F=ma
F=m(g + ay)

The Attempt at a Solution

Part A:

The elevator is initially going in the downward direction. This is because the apparent weight stays the same for 2 seconds if it went upwards the apparent weight would have gone up right then starting from t=0.

Part B:

First I just did F=ma at 600N because that was the start...so 600N=mg. m came out to be 61.22kg. This was incorrect.


Then I thought that whenever there is an apparent weight change that is when the elevator switches direction because it has to break equilibrium in the forces to switch the net forces direction. So I suppose that since the apparent weight is 750N mostly I put that in force and it was: 750N=mg so m= 76.53kg

Part C:

Im not sure how to do this. I suppose there is some relation between apparent weight change and acceleration? like Fnet=ma and the a will be the acceleration for the direction change?

The only thing is the image does not show how long the acceleration is for.

please help for B and C

Can we can agree that if the person is not accelerating, then the apparent weight is just mg?
If the apparent weight is greater than mg, then the floor, or scale on the floor, is pushing back on the person with a force that is greater than mg. This force is mg + ma. If the person has an apparent weight that is less than mg, then his apparent weight (force exerted by the scale on the person) is less than mg, or mg - ma. If you draw a FBD of this and make two situations, net force down and net force up, it is easy to see.

As far as the graph is concerned... If the person's apparent weight changes through time, then the acceleration is changing. Because Weight is a force and mass does not change so acceleration must be the part changing in F = ma. Hope this helps with the concepts.