How Does Elevator Acceleration Affect Object Weight?

[joelio36]

1. Question
A lift has a object of mass 5kg inside. Find the reaction force of the lift upon the object, and the object's relative weight when:
The lift is descending with an acceleration of 3m/s
The lift is descending with constant speed
The lift is decelerating at a rate if 3m/s (while descending)

2. My attempts and reasoning
I am confused about this question. I understand that on Earth the acceleration due to gravity is 9.8m/s/s, and the gravitational field is 9.8n/kg. So when the lift is at constant speed, acceleration is 0, but their is still an acceleration due to gravity (otherwise the object would lift off floor), so i assume weight & reaction = (9.8 * 5)= 49N
When the lift is descending at 3m/s/s, does this mean total acceleration is 9.8 + 3 = 12.8m/s, so weight and reaction both equal 12.8*5= 64N?
When the lift is retarding at 3m/s/s, total acceleration= 9.8-3= 6.8N, therefore weight and reaction both = 6.8 * 5= 34N.

I have no idea if my logic is right, close, or no where near, since this is what my buddy is doing, I haven't done it yet, because we are in different classes. are my answers right, if not, why?

PS- does anyone know a good website to help explain physics concepts at a-level/K10-12, undergrad level?

 

[hunter151]

You have to consider how the forces are acting on the object in this situation.

When the elevator is stationary, there is a gravitational force equal to mg acting downwards. Since the object is stationary, the net force must be zero, thus the upward normal force has to equal the force of gravity (m*g).

When the elevator is accelerating downwards, the object has a total acceleration of 3m/s downward. Thsi means that the normal force is less than the gravitational force, and the net force on the object is Fn + mg = m*(-3). So Fn would be m * (-3 - (-9.8)) taking the downward direction as negative.

For the case when the elevator is decelerating, the same analysis applies, only now the acceleration is 3 m/s upward.