posted by Krystal on .
A 40kg block is pushed along a level surface with a 200 N force at an angle of 35 degrees below the horizon. The coefficient of kinetic friction between the block and the surface is 0.15.
What is the normal force acting on the block?
What is the force of friction?
What is the acceleration of the block?
Wb = mg = 40kg * 9.8N/kg = 392 N. = Wt.
Fb = 392 N. @ 0 Deg.
Fp = 392*sin(0) = 0 = Force parallel to
Fv = 392*cos(0) = 392 N. = Force perpendicular to surface.
a. Fv' = Fv + Fap*sinA,
Fv' = 392 + 200*sin35 = 507 N. = Normal
b. Fk = u*Fv' = 0.15 * 507 = 76 N. =
Force of kinetic friction.
c. Fn = Fap*cos35 - Fp - Fk,
Fn = 200*cos35 - 0 - 76 = 87.8 N. = Net
a = Fn/m = 87.8 / 40 = 2.2 m/s^2.