A 20.0 kg box slides down a 12.0 m long incline at an angle of 30.0 degrees with the horizontal. A force of 50.0 N is applied to the box in the upward direction. The applied force F makes an angle of 0.00 to the incline. If the incline is frictionless, then the increase in the kinetic energy of the box is
A) 620 J.
B) 576 J.
C) 542 J.
D) 500 J.
E) 475 J.

Question

Quizplus · Accepted Answer

The increase in kinetic energy can be calculated using the work-energy principle. The work done on the box is the sum of the work done by gravity and the work done by the applied force. The work done by gravity is $W_g = mgh$, where $h$ is the height of the incline. The height can be found using trigonometry: $h = 12 \sin(30^\circ) = 6.0 \, \text{m}$. Thus, $W_g = (20.0 \, \text{kg})(9.8 \, \text{m/s}^2)(6.0 \, \text{m}) = 1176 \, \text{J}$.The work done by the applied force is $W_f = Fd \cos(\theta)$, where $F = 50.0 \, \text{N}$, $d = 12.0 \, \text{m}$, and $\theta = 0^\circ$, so $W_f = (50.0 \, \text{N})(12.0 \, \text{m})\cos(0^\circ) = 600 \, \text{J}$. However, since the force is applied in the upward direction, opposite to the direction of motion, this work should be subtracted from the work done by gravity, giving a total work (and thus increase in kinetic energy) of $1176 \, \text{J} - 600 \, \text{J} = 576 \, \text{J}$.

A 200 Kg Box Slides Down a 12