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Mary applies a force of 70 N to push a box with an acceleration of 0.50 m/s2. When she increases the pushing force to 78 N, the box's acceleration changes to 0.68 m/s2. There is a constant friction force present between the floor and the box.
(a) What is the mass of the box in kilograms?
(b) What is the coefficient of kinetic friction between the floor and the box?

Respuesta :

Explanation:

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Answer:

The answers are: a) m = 44.44 kg; b) μ = 0.1097

Explanation:

a)

we have the following data:

F = applied force

f = friction force

F - f = net force = ma

F1 = 70 N

a1 = 0.5 m/s^2

F2 = 78 N

a2 = 0.68 m/s^2

F1 - f = ma1

F2 - f = ma2

Subtracting the second equation from the first equation:

F1 - F2 = m(a1 - a2)

m = (F1 - F2)/(a1 - a2) = (70 - 78)/(0.5 - 0.68) = 44.44 kg

b)

f = μmg

F1 - μmg = ma1

Clearing μ:

μ = (F1 - ma1)/(mg) = [70 - (44.44*0.5)]/[44.44*9.8] = 0.1097