Newton's Law on an elevator

1. Sep 24, 2008

crhscoog

1. The problem statement, all variables and given/known data

"A student whose weight is 500 Newtons stands on a scale in an elevator and records the scale reading as a function of time. The data is shown in the graph above. At time t=0, the elevator is at displacement x=0 with velocity v=0. Assume that the positive directions for displacement, velocity, and acceleration are upward.

Graph (scale reading 'newtons' vs time):
0-5 seconds --> 500 Newtons
5-10 seconds --> 700 Newtons
10-15 seconds --> 500 Newtons
15-20 seconds --> 300 Newtons"

3. The attempt at a solution

a) Free Body Diagram w/ all forces on the student at t=8

Normal Force going up
mg going down

b) Calculate the acceleration of the elevator for each 5 second interval

0-5s=
0 m/s^2 because force stays the same meaning its at rest since velocity is 0 at 0seconds

5-10s=
fnet=ma
700-500=51a
a= 3.92m/s^2

10-15s=
force is back to 500 meaning constant velocity
a= 0m/s^2

15-20s=
fnet=ma
300-500=51a
a= -3.92 m/s^2

c) It asks for velocity when t= 5,10,15,20

v=at
v=0(5)
v= 0 m/s

v=vi+at
v=0+3.92(5)
v= 19.6 m/s

v=vi+at
v=19.6+0(5)
v=19.6 m/s
v=19.6+(-3.92)(5)
v=0 m/s

Now I have to plot this on a velocity vs time graph so I have it at:
0 velocity from 0-5s
line from (5,0) to (10,19.6).
horizontal line from (10,19.6) to (15,19.6)
line from (15,19.6) to (20,0)

Just want to see if I am doing this right... Thank you.

2. Sep 24, 2008

Mattowander

Just from a quick skim, it looks like you're doing it correctly.

3. Sep 24, 2008

thank you.