physics
posted by Anonymous on .
A box with a mass of 40.0 kg is pulled along a level floor at a constant speed by a rope that makes an angle of 20.0o with the floor as shown below. If the force on the rope is 200 N, then coefficient of sliding friction or ) between the box and the floor is:
a) 0.58
b) 0.48
c) 0.51
d) 0.30

Fh = 200*cos20 = 187.9N.
Fv = 40*9.8 + 200*sin20,
Fv = 392 + 68.4 = 460.4N.
Since the velocity is constant, the
acceleration = 0:
Fn = ma = 0, Fn = Net force,
Fh  uFv = 0, uFv = force of friction.
187.9  460.4u = 0,
460.4u = 187.9,
u = 187.9 / 460.4 = 0.408.