A box of books weighing 325N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at a angle of 35.2 degrees below the horizontal. Find the coefficient of kinetic friction between the box and the floor

The coefficient of kinetic friction is 0.61

Explanation:A state of constant velocity connotes an acceleration equal to zero {this is evident from Acceleration=(final velocity-initial velocity)/time}

Consequently,a balanced forces system is attained

Force of pulling

F(x)=F×cosα=425N×0.817=347N

is equivalent to the resistance force.

F=U(k)×(Fw+Fsinα)=U(k)×(325N+425N×0,576)=U(k)×570N

Note; U(k) is the coefficient of kinetic friction

Equating the two force components,

The coefficient of static friction ×570N= 347N

The coefficient of static friction = 347N/570N

The coefficient of static friction =0.61

Please note that the coefficient of static friction is dimensionless .

This is as a result of the same unit of forces whose division resulted in the determination of the coefficient of kinetic friction.

graph question answered .. you think better when you are relaxed and happy