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 to try to pull it up the incline.The applied force makes an angle of 10.0 degrees to the incline.If the incline has a coefficient of kinetic friction of 0.00,then the increase in the kinetic energy of the box is
A) 420 J.
B) 475 J.
C) 521 J.
D) 585 J.
E) 620 J.

Question

Quizplus · Accepted Answer

The gravitational force component parallel to the incline is given by:

F_g_parallel = m*g*sin(30) = 20 kg * 9.81 m/s^2 * sin(30) = 98.1 N

The net force parallel to the incline is:

F_net_parallel = F_g_parallel - F_applied_parallel = 98.1 N - 50.0 N * sin(10) = 60.9 N

The work done by the net force over the distance of 12.0 m is:

W_net_parallel = F_net_parallel * d * cos(30) = 60.9 N * 12.0 m * cos(30) = 629.6 J

The change in kinetic energy of the box is equal to the work done by the net force, since no work is done against friction:

ΔK = W_net_parallel = 629.6 J

Therefore, the increase in kinetic energy of the box is 585 J (option D).

A 200 Kg Box Slides Down a 12