Asked by Anon
A cable attached to a block of mass 16.35 kg pulls the block along a horizontal floor at a constant velocity. If the tension in the cable is 9.36 N, what is the coefficient of kinetic friction between the block and the floor?
not sure how to do this one
... I am thinking I use Sum of all Forces = T = ma =F(friction)= -ìkN
not sure how to do this one
... I am thinking I use Sum of all Forces = T = ma =F(friction)= -ìkN
Answers
Answered by
Henry
Wb = mg = 16.36kg * 9.8N/kg = 160.2N =
Wt. of block.
Fp = 160sin(0) = 0 Newtons. = Force
parallel to plane.
Fv = 160cos(0) = 160N. = Force perpen-
dicular to plane.
Fn = Fap - Fp - Ff = ma = 0(a = 0).
9.36 - 0 - Ff = 0,
Ff = uFv = 9.36,
u*160 = 9.36,
u = 0.0585 = Force of friction.
Fn = Net force.
Fap = Force applied.
Wt. of block.
Fp = 160sin(0) = 0 Newtons. = Force
parallel to plane.
Fv = 160cos(0) = 160N. = Force perpen-
dicular to plane.
Fn = Fap - Fp - Ff = ma = 0(a = 0).
9.36 - 0 - Ff = 0,
Ff = uFv = 9.36,
u*160 = 9.36,
u = 0.0585 = Force of friction.
Fn = Net force.
Fap = Force applied.
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