Asked by Emma
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?
...Is this one of those trick questions where the work is 0?
Answers
Answered by
Henry
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.
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.
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