Calculating Elevator Acceleration and Velocity?

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SUMMARY

The discussion focuses on calculating the acceleration and velocity of a student standing on a scale in an elevator, based on recorded scale readings over time. The student weighs 500 Newtons, and the analysis breaks down the elevator's behavior into four 5-second intervals. Key findings include an acceleration of 3.92 m/s² during the interval from 5 to 10 seconds, zero acceleration from 10 to 15 seconds, and a deceleration of -3.92 m/s² from 15 to 20 seconds. The velocity calculations at specified times confirm the changes in motion corresponding to the elevator's acceleration.

PREREQUISITES
  • Understanding of Newton's Second Law (fnet = ma)
  • Knowledge of free body diagrams
  • Familiarity with kinematic equations
  • Basic concepts of acceleration and velocity
NEXT STEPS
  • Study the application of free body diagrams in various physics problems
  • Learn more about kinematic equations and their applications in motion analysis
  • Explore the effects of varying forces on acceleration in different scenarios
  • Investigate the relationship between force, mass, and acceleration in real-world contexts
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of motion in elevators and similar systems.

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Homework Statement



"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"

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.
 
Physics news on Phys.org
Just from a quick skim, it looks like you're doing it correctly.
 
Mattowander said:
Just from a quick skim, it looks like you're doing it correctly.

thank you.
 

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