Question

A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ?=30? above the horizontal, as shown (Figure 6) . The normal force acting on the box has a magnitude n=1.7N, the coefficient of kinetic friction between the box and the plane is ?k=0.30, and the displacement d? of the box is 1.8 m down the inclined plane.
part a.What is the work Wn done on the box by the normal force?

Express your answers in joules to two significant figures.
part b. What is the work Wfk done on the box by the force of kinetic friction?

Answer 1

Answer:

a) 0 J

b) 0.92J

Explanation:

The work is defined as:

$W=F*d*cos(\theta)$

because the normal force is perpendicular to the displacement the work done for it is zero.

$W_N=F_N*d*cos(90)=0$

The force because of frictions has an angle of 180 degrees because it is opposite to the motion, so:

$W_f=F_f*d*cos(180)\\W_f=-\µ*F_N*d\\W_f=-0.30*1.7N*1.8m\\W_f=-0.92J$