Question

An 82.0 N box is pulled 15.0 m up a 30° incline by an applied force of 101 N that points 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.

Answer 1

The box weighs 82pounds, but it is on a 30 degree incline, so the amount which gravity pulls against it [parallel to the incline] is only 82*sin(30)=F_P=41 Newtons

Work done by parallel force = -41*15= -615 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=82*(0.22)*cos(30)

Work done by friction = -82*(0.22)*cos(30)*20 = W_f

[Use calculator to get W_f...should be approximately -233

Work done by applied force = 101*15 = 1515 J

So change in kinetic energy = 1515-233+W_f

So the answer is approximately 695J.