Question

# Mary applies a force of 75 N to push a box with an acceleration of 0.43...

Mary applies a force of 75 N to push a box with an acceleration of 0.43 m/s2. When she increases the pushing force to 81 N, the box's acceleration changes to 0.63 m/s2. There is a constant friction force present between the floor and the box. (a) What is the mass of the box? kg (b) What is the coefficient of kinetic friction between the floor and the box?

(a) What is the mass of the box?
_________kg

(b) What is the coefficient of kinetic friction between the floor and the box?
__________

According to the given problem,

constant force of kinetic friction = f
net force acting on box when it accelerates at 0.43 m/s² = F1 = 75 - f
net force acting on box when it accelerates at 0.63 m/s² = F2 = 81 - f
mass of box = F1/0.43 = (75-f)/0.42
mass of box = F2/0.63 = (81-f)/0.63
(75-f)/0.43 = (81-f)/0.63
0.63(75-f) = 0.43(81-f)
47.25 - 0.63f = 34.82 - 0.43f
0.2f = 12.42
f = 62.1 N
box mass = (75-62.1)/0.43 = 30 kg (a)
coefficient of kinetic friction = f/30g = 62.1/30(9.81) = 0.211 (b)

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