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.
