Asked by Anonymous
An 8kg block starts from rest from the top of a plane, inclined at 40o with respect to the horizontal, and slides down at a
constant acceleration. If the coefficient of kinetic friction between the block the plane is 0.35 determine how far the
block travels in 3s.
2013/
constant acceleration. If the coefficient of kinetic friction between the block the plane is 0.35 determine how far the
block travels in 3s.
2013/
Answers
Answered by
Henry
M*g = 8 * 9.8 = 78.4 N. = Wt. of block.
Fp = 78.4*sin40 = 50.4 N. = Force parallel to the incline.
Fn = 78.4*cos40 = 60.1 N. = Normal force
Fk = u*Fn = 0.35 * 60.1 = 21 N. = Force
of friction.
a = (Fp-Fk)/M = (50.4-21)/8 = 3.68 m/s^2
d = 0.5a*t^2 = 0.5*3.68*3^2 = 16.54 m.
Fp = 78.4*sin40 = 50.4 N. = Force parallel to the incline.
Fn = 78.4*cos40 = 60.1 N. = Normal force
Fk = u*Fn = 0.35 * 60.1 = 21 N. = Force
of friction.
a = (Fp-Fk)/M = (50.4-21)/8 = 3.68 m/s^2
d = 0.5a*t^2 = 0.5*3.68*3^2 = 16.54 m.
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