A 50.0-kg block is being pulled up a 13 degree slope by a force of 250 N which is parallel to the slope. The coefficient of kinetic friction between the block and the slope is 0.200. What is the acceleration of the block?

Answer 1

Wb = mg = 50kg * 9.8N/kg = 490N. = Weight of blocki.

Fb = 490N @ 13deg.
Fp = 490sin13 = 110.2N. = Force parallel to slope.
Fv = 490cos13 = 477.4N. = Force perpendicular to slope.

Fn = Fap - Fp - Ff,
Fn = 250 - 110.2 - 0.2*477.4 = 44.3N. =
Net force.

a = Fn / m = 44.3 / 50 = 0.89m/s^2.