Posted by Krystal on Sunday, February 5, 2012 at 1:02pm.
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?

Physics  Henry, Tuesday, February 7, 2012 at 11:57am
Wb = mg = 40kg * 9.8N/kg = 392 N. = Wt.
of block.
Fb = 392 N. @ 0 Deg.
Fp = 392*sin(0) = 0 = Force parallel to
surface.
Fv = 392*cos(0) = 392 N. = Force perpendicular to surface.
a. Fv' = Fv + Fap*sinA,
Fv' = 392 + 200*sin35 = 507 N. = Normal
force.
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
applied force.
a = Fn/m = 87.8 / 40 = 2.2 m/s^2.
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