Asked by Kathy
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?
Answers
Answered by
Henry
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.
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.
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