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 to try to pull it up the incline.The applied force makes an angle of 10.00 degrees to the incline and above the incline.If the incline has a coefficient of kinetic friction of 0.100,then the increase in the kinetic energy of the box is
A) 203 J.
B) 275 J.
C) 300 J.
D) 392 J.
E) 425 J.

Question

Quizplus · Accepted Answer

First, we need to find the gravitational force acting on the box:
$F_g = mg = (20.0	ext{ kg})(9.81	ext{ m/s}^2) = 196.2$ N

Then, we can find the component of the gravitational force acting parallel to the incline:
$F_{g\parallel} = F_g \sin 	heta = (196.2	ext{ N})\sin 30.0^\circ = 98.1$ N

The frictional force acting against the box can be found using the coefficient of kinetic friction and the normal force:
$F_f = \mu_k F_n = \mu_k m g \cos 	heta = (0.100)(20.0	ext{ kg})(9.81	ext{ m/s}^2)(\cos 30.0^\circ) = 16.9$ N

Note that since the box is sliding down the incline, the frictional force must act up the incline.

Next, we can find the net force acting on the box parallel to the incline:
$F_{net \parallel} = F_{g\parallel} - F_f = (98.1	ext{ N}) - (16.9	ext{ N}) = 81.2$ N

Using the net force, we can find the acceleration of the box:
$a = \frac{F_{net \parallel}}{m} = \frac{81.2	ext{ N}}{20.0	ext{ kg}} = 4.06$ m/s$^2$

The work done by the applied force over the distance of the incline can be found using the component of the force parallel to the incline:
$W_{app} = F_{app} d \cos 	heta' = (50.0	ext{ N})(12.0	ext{ m})(\cos 10.00^\circ) = 596$ J

Again, note that we use the angle of 10.00 degrees since the force is at an angle above the incline.

The increase in the kinetic energy of the box can now be calculated using the work-energy theorem:
$\Delta K = W_{app} - (F_f + F_{g\parallel})d = 596	ext{ J} - ((16.9	ext{ N} + 98.1	ext{ N})	imes 12.0	ext{ m}) = 392$ J

Therefore, the answer is D.

A 200 Kg Box Slides Down a 12