Question

An object with a mass of 13 kg slides 5 meters across a horizontal surface with a coefficient of kinetic friction of 0.38. For the duration of the slide an external force, F1, acts on the mass in the positive y direction while another force, F2, has a magnitude of 67 Newtons and is angled at 28 degrees above the x axis. The net work done over the distance of 5 meters, 499 Joules. What is the magnitude of F1 in Newtons?

Answer 1

Wo = mg = 13kg * 9.8N/kg = 127.4N =
Weight of object.
Fo = 127.4N@odeg. = Force of object.
Fp = 127.4sin(0) = 0 = Force parallel to surface.
Fv = 127.4cos(0) = 127.4N. = Force perpendicular to surface.

W = Fn*d,
Fn = W / d = 499 / 5 = 100N.
Fn = F2cos28 - Fp - Ff = 100.
67cos28 - 0 - Ff = 100,
Ff = 67cos28 - 100 = -40.84N.
Ff = u*Fv = 0.38(127.4+F1) = -40.84,
0.38(127.4 - F1) = -40.84,
48.41 - 0.38F1 = -40.84
-0.38F1 = -40.84 - 48.41 = -89.25,
F1 = 235N.

NOTE: F1 was subtracted from Fv ,
because it acts upward while Fv acts
downward.