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A 6.22-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.412. Determine the kinetic frictional force that acts on the box when the elevator is (a) stationary, (b) accelerating upward with an acceleration whose magnitude is 2.55 m/s2, and (c) accelerating downward with an acceleration whose magnitude is 2.55 m/s2.

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Answer:

a) 25.14 N

b) 31.67 N

c) 18.6 N

Explanation:

m = Mass = 6.22 kg

[tex]\mu[/tex] = Coefficient of friction = 0.412

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

a)

[tex]F_f=\mu mg\\\Rightarrow F_f=0.412\times 6.22\times 9.81\\\Rightarrow F_f=25.14\ N[/tex]

Frictional force = 25.14 N

b) Accelerating up, a = 2.55 m/s²

[tex]F_f=\mu m(a+g)\\\Rightarrow F_f=0.412\times 6.22\times (2.55+9.81)\\\Rightarrow F_f=31.67\ N[/tex]

Frictional force = 31.67 N

c) Accelerating down, a = 2.55 m/s²

[tex]F_f=\mu m(g-a)\\\Rightarrow F_f=0.412\times 6.22\times (9.81-2.55)\\\Rightarrow F_f=18.6\ N[/tex]

Frictional force = 18.6 N

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