Asked by ali
A 40N force applied at an angle of 37 degrees above the horizontal pulls a 5kg box on a horizontal floor. the acceleration of the box is 3meter per second square. how large a frictional force must be retarding the motion of the box.
a.50N b.13N c.17N d.25N
a.50N b.13N c.17N d.25N
Answers
Answered by
Henry
Wb = mg = 5kg * 9.8N/kg = 49 N.
Fb = 49N @ 0 Deg. = Force of box.
Fp = 49*sin(0) = 0. = Force parallel to
floor.
Fv = 49*cos(0) = 49 N. = Force perpendicular to floor.
Fv'=Fv-Fap*sin37 = 49-40sin37 = 24.9 N.
= Normal. = Net force perpendicular to
floor.
Fn = Fap*cosA-Fp-Fk = ma.
40*cos37-0-Fk = 5*3.
31.9 - Fk = 15.
Fk = 31.9 - 15 = 17.N. = Force of kinetic friction.
Fb = 49N @ 0 Deg. = Force of box.
Fp = 49*sin(0) = 0. = Force parallel to
floor.
Fv = 49*cos(0) = 49 N. = Force perpendicular to floor.
Fv'=Fv-Fap*sin37 = 49-40sin37 = 24.9 N.
= Normal. = Net force perpendicular to
floor.
Fn = Fap*cosA-Fp-Fk = ma.
40*cos37-0-Fk = 5*3.
31.9 - Fk = 15.
Fk = 31.9 - 15 = 17.N. = Force of kinetic friction.
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