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?

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

newton's first law of motion

the part of the wave he is travelling on is the crest