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Nookie1986 [14]
2 years ago
15

A force of 36.0 N is required to start a 3.0-kg box moving across a horizontal concrete floor. Part A) What is the coefficient o

f static friction between the box and the floor? Express your answer using two significant figures. Part B) If the 36.0-N force continues, the box accelerates at 0.54 m/s^2. What is the coefficient of kinetic friction? Express your answer using two significant figures.
Physics
1 answer:
matrenka [14]2 years ago
4 0

Answer:

A) 1.2

B) 1.1

Explanation:

A)

F = force required to start the box moving = 36.0 N

m = mass of the box = 3 kg

F_{g}  = weight of the box = mg = 3 x 9.8 = 29.4 N

F_{n}  = normal force acting on the box by the floor

normal force acting on the box by the floor is given as

F_{n}  = F_{g} = 29.4

F_{s}  = Static frictional force = F = 36.0 N

\mu _{s} = Coefficient of static friction

Static frictional force is given as

F_{s}  = \mu _{s} F_{n}

36.0  = \mu _{s} (29.4)

\mu _{s} = 1.2

B)

a = acceleration of the box = 0.54 m/s²

F = force applied = 36.0 N

f_{k} = kinetic frictional force

\mu _{k} = Coefficient of kinetic friction

force equation for the motion of the box is given as

F - f_{k} = ma

36.0 - f_{k} = (3) (0.54)

f_{k} = 34.38 N

Coefficient of kinetic friction is given as

\mu _{k}=\frac{f_{k}}{F_{n}}

\mu _{k}=\frac{34.38}{29.4}

\mu _{k} = 1.1

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Answer:

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Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

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We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

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