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ankoles [38]
3 years ago
5

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?

Physics
2 answers:
qwelly [4]3 years ago
6 0

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

neonofarm [45]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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