Apparent Weight of 59.3kg Student in Elevator Forces

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The discussion revolves around calculating the apparent weight of a 59.3 kg student in an elevator under various conditions using basic physics equations. For constant velocity upwards, the apparent weight remains equal to the gravitational force, while during constant acceleration downwards, it decreases due to the opposing force. Conversely, with constant acceleration upwards, the apparent weight increases, and during freefall, the scale reads zero. Participants emphasize the importance of distinguishing between weight (in Newtons) and mass (in kilograms) when applying the force equations. The conversation highlights the necessity of correctly applying the formula Fnet = ma to derive accurate results for each scenario.
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Intro: A physics student of a mass 59.3kg is standing on a scale in an elecator. The scale shows the magnitude of the upward normal force (in Newtons) on the student, this value is called the apparent weight. What is the reading on the scale when the elevator:

Questions

a) has a constant velocity of 3.56 m/s [up] ?
b) has a constant acceleration of 3.56 m/s squared [down] ?
c) has a constant acceleration of 3.56 m/s squared [up] ?
d) is undergoing freefall?

Answer using simple Force equations, such as:
Fnet = ma = Fg + Fn + Fapp

a = Fnet
m

and other easy (grade 12) type equations. Thanks
 
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You need to show some work on your own before we can help you. What are your answers to a-d so far?
 
im not sure if I am over thinking this or not, but i tried to think simple and i came up with...
a) 59.3kg
b)42.6kg
c)76.0kg
d)0kg
 
You're on the right track, but remember that weight is the force F produced on a mass m by the acceleration of gravity g. What is the equation relating force, mass and acceleration?
 
Fnet = ma ?
 
so would b) = 37.8 kg ?
 
and a) = 80.8 kg ?
 
sorry not a)...i mean c)
 
No. Weight has units of Newtons (N), and mass has units of kg. Let's take the stationary example:

F = mg = 59.3 kg * 9.8 \frac{m}{s^2} = ? N
 

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