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An 80 N box is pulled 20 m up a 30o Incline by an applied force of 100N that point upward, parallel to the incline. If the coefficient of kinetic friction between box and incline is 0.220, calculate the change in the kinetic energy of the box.

2007-10-27 12:26:59 · 1 answers · asked by info2know 3 in Science & Mathematics Physics

1 answers

General idea: Change in kinetic energy = Work done by the force plus work done by gravity plus work done by friction. [Note these latter two will be negative because the force is in the opposite direction as the movement].

The box weighs 80 pounds, but it is on a 30 degree incline, so the amount which gravity pulls against it [parallel to the incline] is only 80*sin(30)=F_P=40 Newtons

Work done by parallel force = -40*20 = -800 Joules.

The box pushes into the incline with a force equal to the perpendicular gravitational force: 80*cos(30). The force of friction is equal to this times the coefficient of friction, so

F_f=80*(0.22)*cos(30)

Work done by friction = -80*(0.22)*cos(30)*20 = W_f

[Use calculator to get W_f...should be approximately -280

Work done by applied force = 100*20 = 2000 J

So change in kinetic energy = 2000-800+W_f

So the answer is approximately 920 J.

2007-10-27 13:42:59 · answer #1 · answered by David Zukertort Rudel 3 · 0 0

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