Solving the Elevator Problem: Weight, Speed & Acceleration

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To determine the scale reading for the girl in the elevator, first calculate the acceleration, which is negative due to the elevator's deceleration. The acceleration can be found using the formula a = Δv/Δt, where the change in velocity is -4 m/s and the time interval is 1 second. Using Newton's second law, the effective weight on the scale is the sum of gravitational force and the force due to the elevator's acceleration. The scale will read less than her actual weight because the elevator is slowing down while ascending. Understanding the direction of acceleration is crucial for accurately calculating the forces involved.
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A 56-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when the elevator is ascending at 13m/s buts its speed is decreasing by 4m/s in each second?

I know you use Newton's second law, however, I do not know how to find acceleration..
 
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which direction is the acceration? If you figure that out you know which direction the force of the slowing elevator is. Add the wieght vector with the force vector to get her wieght...




the acceration if opposite the velocity ( girl weighs W = mg + ma_elevator)
 
khf said:
I know you use Newton's second law, however, I do not know how to find acceleration..
Apply the definition of acceleration: a = \Delta v/\Delta t. (Direction matters.)
 

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