Question

You want to move a 4- kg bookcase to a different place in the living room. If u push with a force of 65 n and the bookcase accelerates at 0.12 m/s2 what is the coefficient of kinetic friction between the bookcase and the carpet

Answer 1

1.65

Step-by-step explanation:

The equation of the forces along the horizontal direction is:

`F-F_f = ma` (1)

where

F = 65 N is the force applied with the push

`F_f` is the frictional force

m = 4 kg is the mass

`a=0.12 m/s^2` is the acceleration

The force of friction can be written as
`F_f = \mu R` (2), where

`\mu` is the coefficient of kinetic friction

R is the normal force exerted by the floor

The equation of forces along the vertical direction is

`R-mg=0` (3)

since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find

`F-\mu mg = ma`

And solving for
`\mu`,

`\mu = (F-ma)/(mg)=(65-(4)(0.12))/(4(9.8))=1.65`