Asked by christoper allen
a force of 200 N is exerted at an angle 30 degrees with respect to the horizontal on an object whose mass is 30kg. the coefficient of friction between the object and the surface is the 0.2.(a)what is the frictional force?(B) what is the acceleration of the object?(c) if the object starts from rest what is the velocity after after 5 seconds ?
Answers
Answered by
Henry
F = 200N[30o]
Fx = 200*Cos30 = 173.2 N.
Fy = 200*sin30 = 100 N.
M*g = 30kg * 9.8N/kg = 294 N. = Wt. of
the object.
a. Fk = u*Fn = u*(Mg-Fy) =
0.2(294-100) = 38.8 N. = Force of kinetic friction.
b. a = (Fx-Fk)/M
c. V = Vo + a*t
Vo = 0
a = Value from part b.
t = 5 s.
Solve for V.
Fx = 200*Cos30 = 173.2 N.
Fy = 200*sin30 = 100 N.
M*g = 30kg * 9.8N/kg = 294 N. = Wt. of
the object.
a. Fk = u*Fn = u*(Mg-Fy) =
0.2(294-100) = 38.8 N. = Force of kinetic friction.
b. a = (Fx-Fk)/M
c. V = Vo + a*t
Vo = 0
a = Value from part b.
t = 5 s.
Solve for V.
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