Asked by Roman
A 25 kg box rests on an inclined plane which is 30o to the horizontal. The coefficient of sliding friction between the box and the plane is 0.30. Determine the acceleration of the box down the plane. (9.8 m/s2)
Answers
Answered by
Henry
M*g = 25 * 9.8 = 245 N. = Wt. of the box
Fp = 245*sin30 = 122.5 N. = Force parallel to the incline.
Fn = 245*Cos30 = 212.2 N. = Force perpendicular to the incline.
Fk = u*Fn = 0.30 * 212.2 N. = 63.65 N. = Force of kinetic friction.
a = (Fp-Fk)/M*sin30 = (122.5-63.65)/12.5
= 4.71 m/s^2
Fp = 245*sin30 = 122.5 N. = Force parallel to the incline.
Fn = 245*Cos30 = 212.2 N. = Force perpendicular to the incline.
Fk = u*Fn = 0.30 * 212.2 N. = 63.65 N. = Force of kinetic friction.
a = (Fp-Fk)/M*sin30 = (122.5-63.65)/12.5
= 4.71 m/s^2
Answered by
Henry
Correction:
a = (Fp-Fk)/M = (122.6-63.65)/25 = 2.36
m/s^2
a = (Fp-Fk)/M = (122.6-63.65)/25 = 2.36
m/s^2
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