Asked by Imani
A block of mass m = 5.9 kg is pulled up a è = 21° incline as in the figure with a force of magnitude F = 34 N.
(a) Find the acceleration of the block if the incline is frictionless.
1
Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2
(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12.
(a) Find the acceleration of the block if the incline is frictionless.
1
Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2
(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12.
Answers
Answered by
Henry
Wb = mg = 5.9kg * 9.8N/kg = 57.82N. =
Weight of block.
Fb = (57.82N,21deg.) = Force of block.
Fp = 57.82sin21 = 20.72N. = Force parallel to the incline.
Fv = 57.82cos21 = 53.98N. = Force perpendicular to incline. = Tne Normal.
Fn = Fap-Fp = 34 - 20.72 = 13.28N. =
Net force. Fap = Force applied.
a. Fn = ma,
a = Fn/m = 13.28 / 5.9 = 2.25m/s^2.
b. Ff = u*Fv = 0.12 * 53.98 = 6.48N. =
Force of friction.
Fn = Fap - Fp -Ff = 34 - 20.72 - 6.48 =
6.8N.
a = 6.8 / 5.9 = 1.15m/s^2.
Weight of block.
Fb = (57.82N,21deg.) = Force of block.
Fp = 57.82sin21 = 20.72N. = Force parallel to the incline.
Fv = 57.82cos21 = 53.98N. = Force perpendicular to incline. = Tne Normal.
Fn = Fap-Fp = 34 - 20.72 = 13.28N. =
Net force. Fap = Force applied.
a. Fn = ma,
a = Fn/m = 13.28 / 5.9 = 2.25m/s^2.
b. Ff = u*Fv = 0.12 * 53.98 = 6.48N. =
Force of friction.
Fn = Fap - Fp -Ff = 34 - 20.72 - 6.48 =
6.8N.
a = 6.8 / 5.9 = 1.15m/s^2.
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