Wb = mg = 28.0kg * 9.8N/kg = 274.4N. =
Weight of block.
W2 = mg = 2.20kg * 9.8N/kg = 21.56N. =
Weight of bucket.
Fb = 274.4N @ 0deg.,
Fp = 274.4sin(0) = 0 = Force parallel to table.
Fv = 274.4cos(0) = 274.4N. = Force
perpendicular to table.
Ff = u*Fv = 0.44 * 274.4 = 120.7N =
Force of static friction.
Ff' = 0.35 * 274.4 = 96.04N. = Force of kinetic friction.
1. Fn=(Fap+F2) - Fp - Ff = 0 = Net force,
(Fap+21.56) - 0 - 120.7 = 0,
Fap = 120.7 - 21.56 = 99.14N. = Force
applied = Force added to bucket.
Fap = mg = 99.14N.,
9.8m = 99.14,
m = 10.1kg = mass added to the bucket.
2. Fn = Fap + F2 - Ff',
Fn = 99.14 + 21.56 - 96.04 = 24.66N.
a = (Fp-Fap) / (m1+m2),
a = (0-99.14 / (28+12.3) = -2.46m/s.
A 28.0 block is connected to an empty 2.20 bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.44 and the coefficient of kinetic friction between the table and the block is 0.35. Sand is gradually added to the bucket until the system just begins to move .Calculate the mass of sand added to the bucket.Calculate the acceleration of the system.
2 answers
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