Question

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?

Answer 1

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.

Answer 2

The force of static friction b/n abody of mass 40kg & horizontal surface is measured to be 200N what is the coefficent of friction b/n the box & the road