A crate with a total mass of 18.0 kg is pulled by a person along the floor at a constant speed by a rope. The rope is inclined at 20 degree above the horizontal, and the crate moves 20.0 m on a horizontal surface. The coefficient of kinetic friction between the floor and the crate is 0.50

Answer 1

And the questions are as follows:
a. Draw a free body diagram for the crate (solved)
b. What is the tension in the rope?
c. How much work is done on the crate by the rope?
d. What is the energy lost due to friction?

Answer 2

Wt. of crate=m*g=18kg * 9.8N/kg=176.4 N.

Fv = 176.4*cos0 = 176.4 N. = Force perpendicular to the surface.

Fk = u*Fv = 0.5 * 176.4 = 88.2 N. = Force of kinetic friction.

Fap*cos20-Fk = m*a
Fap*0.94-88.2 = 18*0 = 0
0.94Fap = 88.2
Fap = 93.9 N. = Force applied

b. Tension = Fap = 93.9 N.

c. Work=Fap*cos20o*d=93.9cos20 * 20=1764 J.

d. E = Fk*d = 88.2 * 20 = 1764 J.

Answer 3

Thanks!