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Today, 10:51

You are riding in an elevator on the way to the

eighteenthfloor of your dormitory. The elevator is accelerating

upwardwith a = 1.90 m/s2. Beside you is the

boxcontaining your new computer; box and contents have a total

mass28.0 kg. While the elevator is accelerating upward, you

pushhorizontally on the box to slide it at constant speed toward

thedoor. If the coefficient of kinetic friction between the boxand

elevator floor is mk = 0.32, what magnitudeof

force must you apply?

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Answers (1)
  1. Today, 11:50
    0
    F = 104.832 N

    Explanation:

    given,

    upward acceleration of the lift = 1.90 m/s²

    mass of box containing new computer = 28 kg.

    coefficient of friction = 0.32

    magnitude of force = ?

    box is moving at constant speed hence acceleration will be zero.

    Now force acting due to lift moving upward =

    F = μ m (g + a)

    F = 0.32 * 28 * (9.8 + 1.9)

    F = 104.832 N

    hence, the force applied should be equal to 104.832 N
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