The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.
A)42 N, due west
B)120 N, due east
C)31 N, due east
D)59 N, due east
E)83 N, due west

Question

Quizplus · Accepted Answer

We need to use the work-energy principle to solve this problem. The net work done on an object is equal to its change in kinetic energy. We can split the net work into two parts: the work done by conservative forces and the work done by non-conservative forces. If the non-conservative force is friction, then it is given by:

f_nc = (1/2)m(v_f^2 - v_i^2)/d

where m is the mass of the box, v_f is the final velocity of the box, v_i is the initial velocity of the box, and d is the distance the box slides.

Plugging in the given values, we get:

f_nc = (1/2)(4.0 kg)((1.5 m/s)^2 - (11 m/s)^2)/(4.0 m) = -59 N

The negative sign indicates that the force is opposite to the direction of motion (east). Therefore, the magnitude of the force is 59 N, and the direction is east.

The correct answer is D) 59 N, due east.

The Initial Velocity of a 4