Elevator Problem w/ app. Weight

[Bachi234]

Homework Statement

Henry gets into an elevator on the 50th floor of a building and it begins moving at t = 0 s. The figure shows his apparent weight over the next 12 s where w1 = 520, w2 = 720, and w3 = 920.
The graph:
Begins at t=0 at w1 and goes in a horizontal straight line until 2. At t =2 the line goes straight up to w2 and stays horizontal on w2 until t=10. At t=10 the line goes straight up to w3 and stays horizontal on w3 until t=12. At t=12 it goes back down to w2. That is the end of the graph.

1. Is the elevator's initial direction up or down and how?
2. What is Henry's mass? _____kg
3. How far did Henry travel at t=12 sec?_____m

Homework Equations

F = ma
Wapp = m(g + ay)
Fsp = w + may
X = Vot + .5a(t)^2

The Attempt at a Solution

Number 1- I got that it is going down but I don't understand why. I think it has something to do with the acceleration being positive. What is a clear explanation for why its inital direction is down?

Number 2- I've been trying to use the apparent weight formula to figure out the mass but I can't do that without the acceleration. After working the equations out using the Wapp and the g, I got that " m = (520 / (9.8 + ay)) " But, I can't fully solve for m without the acceleration and I can't solve for acceleration without the total force or m. I seem to be in a hole.

Number 3- Well, after I get the acceleration I can use time and Vo (which would equal 0, right?) to get the distance traveled.

Thanks in advance for your help!

 

[Mattowander]

For number 1, it's actually going up. Picture the elevator in your mind. If it's going up, then it will be pushing up against you harder than usual right?

For number 2, all you need to solve for his mass is his "true weight". This is given to you in the problem if you look closely. F = ma.

 

[baba89]

1. The elevator's initial direction is down because the graph shows that the apparent weight decreases from w1 to w2, indicating that the elevator is moving in a downward direction. This is also supported by the fact that the graph starts at the highest apparent weight (w1) and goes down to the lowest apparent weight (w2).

2. To determine Henry's mass, we can use the formula Wapp = m(g + ay), where Wapp is the apparent weight, g is the acceleration due to gravity (9.8 m/s^2), and ay is the acceleration in the y-direction. We can rearrange this equation to solve for m, which gives us m = Wapp / (g + ay). Since we know that Wapp = 520 N at t=0 and Wapp = 720 N at t=2, we can plug these values in to get m = 520 N / (9.8 m/s^2 + ay) and m = 720 N / (9.8 m/s^2 + ay). We can then set these two equations equal to each other and solve for ay, which will give us the acceleration in the y-direction. Once we have the acceleration, we can plug it back into either of the equations to solve for m.

3. To find the distance traveled at t=12 seconds, we can use the equation X = Vot + 0.5at^2, where X is the distance traveled, Vo is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time. Since we already solved for the acceleration in the previous question, we can plug that value in along with t=12 seconds to calculate the distance traveled.