Weight in N with change in acceleration

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SUMMARY

The discussion centers on the calculation of tension in a scale when a 2kg weight is suspended in an elevator that decelerates uniformly from a constant velocity of 10 m/s to rest over 4 seconds. The correct formula for final velocity is Vf = Vi + at, which was initially misapplied as Vf = Vi*t + 1/2a*t^2. The accurate calculation shows that the tension in the scale reads 14.6N during the deceleration phase, contrasting with the incorrect value of 29.6N derived from the misapplication of kinematic equations.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with kinematic equations
  • Knowledge of gravitational force calculations
  • Basic principles of motion in an accelerating frame of reference
NEXT STEPS
  • Study the correct application of kinematic equations in varying frames of reference
  • Learn about tension forces in non-inertial frames
  • Explore the effects of acceleration on weight measurement
  • Review examples of Newton's laws in elevator scenarios
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of forces in accelerating systems, particularly in relation to tension measurements in elevators.

lajekahr
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2kg weight is suspended on a tension scale that measures in Newtons.
The elevator is moving at constant velocity up of 10m/s. Its velocity is then uniformly reduced to zero in 4 seconds. So that it comes to rest in 4 seconds. What does the scale read during those 4 seconds.

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Using frame of reference with positive down: (That way gravity's acceleration is positive.)
Using Kinematics:
Vf=Vi*t+1/2a*t^2
Therefore:
a=(-2Vi)/t=-5m/s^2
Applying F=ma:
Ftension+Fgravity=ma
-Ft+mg=ma
Ft=mg-ma=2*9.8-2*(-5)=29.6N

My books says: 14.6N

What'd I do wrong?
 
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lajekahr said:
2kg weight is suspended on a tension scale that measures in Newtons.
The elevator is moving at constant velocity up of 10m/s. Its velocity is then uniformly reduced to zero in 4 seconds. So that it comes to rest in 4 seconds. What does the scale read during those 4 seconds.

---------
Using frame of reference with positive down: (That way gravity's acceleration is positive.)
Using Kinematics:
Vf=Vi*t+1/2a*t^2
Therefore:
a=(-2Vi)/t=-5m/s^2
Applying F=ma:
Ftension+Fgravity=ma
-Ft+mg=ma
Ft=mg-ma=2*9.8-2*(-5)=29.6N

My books says: 14.6N

What'd I do wrong?
Your first equation is wrong.
It should be Vf=Vi+at.
 
Now i feel kinda dumb. I knew that I swear...that first equation is for displacement...doh!
 

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