A 5.3kg block is released from the top of a 20degree incline. If the coefficent of kinetic friction is .33, what is the acceleration of the block down the incline?
Wb = mg = 5.3 kg * 9.8 N./kg = 51.94 N.
= Weight of block.
Fb = 51.94 N. @ 20 Deg. = Force of block.
Fp = 51.94*sin20 = 17.76 N. = Force
parallel to plane.
Fv = 51.94*cos20 = 48.80 N. = Force perpendicular to plane.
Fk = u*Fv = 0.33*48.80 = 16.11 N. =
Force of kinetic friction.
Fn = Fp-Fk = ma,
17.76-16.11 = 5.3a,
1.653 = 5.3a,
a = 0.31 m/s^2.