A 2.60-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.380. Determine the kinetic frictional force that acts on the box for each of the following cases. (a) The elevator is stationary. N (b) The elevator is accelerating upward with an acceleration whose magnitude is 1.20 m/s2. N (c) The elevator is accelerating downward with an acceleration whose magnitude is 1.20 m/s2. N Additional Materials

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whitneytr12 whitneytr12 · Answer 1 · 2020-10-07T02:36:37+02:00

Answer:

a) F = 9.69 N

b) F = 10.88 N

c) F = 8.51 N

Explanation:

a) The kinetic frictional force when the elevator is stationary is the following:

[tex] F_{k} = \mu_{k}N = \mu(mg) [/tex]

Where:

F(k) is the kinetic frictional force

N is the normal force = mg

m: is the mass = 2.60 kg

g: is the gravity = 9.81 m/s²

μ(k) is the coefficient of kinetic friction = 0.380

[tex] F_{k} = \mu(mg) = 0.380*2.60 kg*9.81 m/s^{2} = 9.69 N [/tex]

b) When the elevator is accelerating upward with acceleration "a" equal to 1.20 m/s²

[tex] F_{k} = \mu_{k}N = \mu[m(g + a)] [/tex]

The normal is equal to mg plus ma because the elevator is accelerating upward

[tex] F_{k} = \mu m(g + a) = 0.380*2.60 kg(9.81 m/s^{2} + 1.20 m/s^{2}) = 10.88 N [/tex]

c) When the elevator is accelerating downward with a =1.20 m/s² we can find the kinetic frictional force similar to the previous case:

[tex] F_{k} = \mu m(g - a) = 0.380*2.60 kg(9.81 m/s^{2} - 1.20 m/s^{2}) = 8.51 N [/tex]

I hope it helps you!