Asked by Edward
15kg block rests on a rough surface (u=.35) a 100N force to the left is applied to the block.
what is the magnitude and direction of the net force acting on the block?
How quickly will the block accelerate over the rough surface?
what is the magnitude and direction of the net force acting on the block?
How quickly will the block accelerate over the rough surface?
Answers
Answered by
Henry
1. Fb = mg = 15kg * 9.8N/kg = 147N.
147N @ 0 deg.
Fp = 147sin(0) = 0 = Force parallel to plane(hor.).
Fv = 147cos(0) = 147N = Force perpendicular to plane.
Ff = u*Fv = 0.35 * 147 = 51.45N.
Fn = Fap - Fp - Ff,
Fn = 100 - 0 - 51.45 = 48.6N = Net force parallel to the hor plane.
147N @ 0 deg.
Fp = 147sin(0) = 0 = Force parallel to plane(hor.).
Fv = 147cos(0) = 147N = Force perpendicular to plane.
Ff = u*Fv = 0.35 * 147 = 51.45N.
Fn = Fap - Fp - Ff,
Fn = 100 - 0 - 51.45 = 48.6N = Net force parallel to the hor plane.
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