Question

1. A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.50 m/s by pulling on the rope tied to the crate with a force of 50 N. The rope makes an angle of 25¡ã with the horizontal.
a.What is the magnitude of each of the forces?
b.How much work is done by each of the forces?
c.What is the total amount of work done on the object?
d.What is the coefficient of friction of the crate on the floor? (Note, we usually write coefficients of friction to only 2 digits of accuracy.)

Answer 1

Wc = mg = 20kg * 9.8N/kg=196N.= Weight of crate.

a. Fc = [196N,0deg).
Fp = 196sin(0) = 0 = Force parallel to
floor.
Fv = 196cos(0) = 196N. = Force perpendicular to floor.

Fn = Fap*cosA - Fp - Ff = 0,
Fn = 50cos25 - 0 - Ff = 0,
50cos25 - Ff = 0,
Ff = 50cos25 = 45.32N. = Force of friction.

b. W = Fd.
W = 50*cosw25 * 12 = 544N.
W = Fp*d = 0 * 12 = 0.
W = Ff*d = 45.32 * 12 = 544N.

c. Wt = 544 + 544 = 1088N.

d. u = Ff / Fv = 45.32 / 196 = 0.23.

Answer 2

Correction: All work is in Joules and NOT Newtons.