Wb = M*g = 3.1 * 9.8 = 30.4 N.
Fp = 30.4*sin28 = 14.3 N. = Force parallel to the incline.
Fn = 30.4*Cos28 = 26.8 N. = Normal force
Fp-Fk = M*a.
Fp-Fk = M*0 = 0.
Fk = Fp = 14.3 N. = Force of kinetic
friction.
u = Fk/Fn = 14.3/26.8 = 0.533.
The 3.1 kg box shown below slides with a constant speed down an incline at an angle of θ = 28.0° to the horizontal.
What is the coefficient of kinetic friction?
1 answer