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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