Scale in an elevator, time, mass, and velocity function given

AI Thread Summary
The discussion centers on calculating the apparent weight of a person standing on a scale in an elevator, using the provided velocity function v(t) = (3 m/s²)t + (0.2 m/s³)t². The acceleration is derived as A(t) = 0.4t + 3, leading to an acceleration of 4.6 m/s² at t=4 seconds. The apparent weight is determined by the normal force, which combines the gravitational force with the force due to acceleration. The correct calculation shows that the normal force is approximately 922 N, confirming the answer. The problem illustrates the relationship between mass, acceleration, and apparent weight in a non-static environment.
Brandan
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Homework Statement


You are standing on a bathroom scale in an elevator in a small building. The elevator starts from rest and moves with a velocity given by v(t) = (3 m/s^2)t + (0.2 m/s^3)t^2. If your mass is 64 kg what is your weight at t=4.0 s as recorded by the scale (in N)

Homework Equations


v(t) = (3 m/s^2)t + (0.2 m/s^3)t^2
The acceleration function is the derivative of the acceleration function so I guess: A(t) = .4t + 3
I think this is needed: F = (ma)

The Attempt at a Solution


So I plug t=4 into the acceleration equation A(t) = .4t + 3 and get 4.6. Then F = m*a so 64*4.6 but that gives me 294N but the answers in the question are 627N, 1254N, 333N, 124N and 922N (the correct answer is 922N btw)

I would really appreciate any help on this problem.
 
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Brandan said:

Homework Statement


You are standing on a bathroom scale in an elevator in a small building. The elevator starts from rest and moves with a velocity given by v(t) = (3 m/s^2)t + (0.2 m/s^3)t^2. If your mass is 64 kg what is your weight at t=4.0 s as recorded by the scale (in N)

Homework Equations


v(t) = (3 m/s^2)t + (0.2 m/s^3)t^2
The acceleration function is the derivative of the acceleration function so I guess: A(t) = .4t + 3
I think this is needed: F = (ma)

The Attempt at a Solution


So I plug t=4 into the acceleration equation A(t) = .4t + 3 and get 4.6. Then F = m*a so 64*4.6 but that gives me 294N but the answers in the question are 627N, 1254N, 333N, 124N and 922N (the correct answer is 922N btw)

I would really appreciate any help on this problem.
You stand on the scales, that exerts a normal force on you, upward. The scale reads the normal force (divided by g). At the same time, gravity pulls downward with force mg. Ma= the sum of forces, N-mg. You accelerate upward, with acceleration 4.6 m/s2. Your apparent weight is equal to the normal force. How much is the normal force?
 
294N + (64kg * 9.8m/s^2) = 921.2 N, thanks I got it
 
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