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Posted by 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 - , 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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