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

Answer:

1.65

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 $\mu$,

$\mu = \frac{F-ma}{mg}=\frac{65-(4)(0.12)}{4(9.8)}=1.65$