Question

A mover pushes a 60 lb box 30 ft along a level floor at constant speed with a force directed 45 degrees below the horizontal. If the coefficient of kinetic friction is 0.20, how much work does the mover do on the box?

...Is this one of those trick questions where the work is 0?

Answer 1

Mass = 60Lbs * 0.454kg/Lb = 27.24 kg

M*g = 27.24 * 9.8 = 267 N. = Wt. of box

Fn = 267 + Fap*sin45 = Normal force.

Fk = u*Fn = 0.2*(267+Fap*sin45) =
53.4 + 0.141Fap

Fap*Cos45-Fk = M*a
0.707Fap-53.4 - 0.141Fap = M*0 = 0
0.566Fap = 53.4
Fap = 94.3 N. = Force applied.

d = 30/3.3 = 9.09 m.

Work = Fap*Cos45 * d = 94.3*Cos45 * 9.09
= 606 J.