A 64kg person is standing on a scale in an elevator which moves down at a constant velocity. It eventually slows and comes to a stop. ?

What is the reading on the scale when while the elevator is accelerating?

The magnitude of the elevator's acceleration is 0.73m/s^2.

2 Answers

  • 1 month ago
    Favorite Answer

    Newton's Second Law:

    F = ma

    Applied to the person's weight:

    W = mg

    W = (64 kg)(9.81 m/s²)

    W = 628 N = 141 lbs

    Now add the acceleration of the elevator:

    W = (64)(9.81 + 0.73)

    W = 675 N = 152 lbs

  • 1 month ago

    from your first sentence, the elevator is de-accelerating 

    scale reads weight plus force needed to de-accelerate

    weight is 64x9.8 newtons

    force = ma = 64x0.73 newtons

    total is the sum of those two.

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