Question

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 of 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.

Answer 1

A) 1.2

B) 1.1

Step-by-step 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)=(f_(k))/(F_(n))`

`\mu _(k)=(34.38)/(29.4)`

`\mu _(k)` = 1.1