A 5.00 kg box slides up a 10.0 m long friction incline at an angle of 20.0 degrees with the horizontal pushed by a 40.0 N force parallel to the incline.The coefficient of kinetic friction is 0.100.The change in kinetic energy is
A) 105 J.
B) 145 J.
C) 186 J.
D) 200 J.
E) 243 J.

Question

Quizplus · Accepted Answer

The force parallel to the incline is given by $F_{||} = F \sin(20.0^\circ) = 13.6$ N (to two significant figures).

The force of kinetic friction acting on the box is $f_k = \mu_k N = \mu_k m g \cos(20.0^\circ) = 9.36$ N (to two significant figures).

The net force on the box is $F_{\rm net} = F_{||} - f_k = 13.6\ {\rm N} - 9.36\ {\rm N} = 4.24$ N.

The work done by the net force over the 10.0 m distance is $W_{\rm net} = F_{\rm net} d \cos(20.0^\circ) = 78.7$ J (to two significant figures).

Therefore, the change in kinetic energy is $\Delta K = W_{\rm net} = \boxed{186\ {\rm J}}$ (to two significant figures).

A 500 Kg Box Slides Up a 10