Ek physics 1001 elevator problem 289

[GRod18]

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Refer to the elevator shown below (basic elevator with cable on top) The mass of the elevator and everything in it is 1000 Kg. The man in the elevator has a mass of 90 kg. The box that he holds has a mass of 10 kg. The floors of the building in which the elevator operates are separated by 5 m. The elevator accelerates and decelerates at 1 m/s^2, and moves at a maximum velocity of 5m/s. The scale reads in newtons.

What is the maximum apparent weight of the box during a trip from 3rd floor to the 1st floor?
Answer: 110 Newtons.

EK: the apparent weight of the box is the force necessary to hold it up. The maximum will occur when the elevator accelerates upward. W = mg + ma.
but from 3rd floor to 1st floor are we not accelerating downwards?


I appreciate everyone that has helped me with these EK problems, EK has about the worst explanation for their physical sciences sections, and I thank everyone who has taken the time to go over these problems with me. 👍

 

[fizzgig]

the apparent weight you see when decelerating the elevator when it falls will yield the same equation. if y is positive in the downward direction, mg points down, ma points up:

F = m(10) - m(-1)

 

[GRod18]

the apparent weight you see when decelerating the elevator when it falls will yield the same equation. if y is positive in the downward direction, mg points down, ma points up:

F = m(10) - m(-1)

ah ic, thanks for clearing that up!