Question

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

Answer 1

Answer: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.