Starting from rest, a 5.6-kg block slides 2.60 m down a rough 30.0° incline.

Starting from rest, a 5.6-kg block slides 2.60 m down a rough 30.0° incline. The coefficient of kinetic friction between the block and the incline is µk = 0.436.

(a) Determine the work done by the force of gravity.
J

(b) Determine the work done by the friction force between block and incline.
J

(c) Determine the work done by the normal force.
J

1st Step => sketch the situation showing all forces that act on block including the two components of block's weight (mg), one that is parallel and one that is perpendicular to the incline.

2nd Step => calculate the work done by the component of weight parallel to plane
Wp = mg sin 30 = (5.6)(9.8)(0.5) = 27.44 N
work done by 27.44 force in moving block 2.60 m = (27.44)(2.60) = 71 J ANS (a)

3rd Step => calculate force of kinetic friction = Kf
Kf = Normal force x 0.436
Normal force = Wn = mg cos 30 = (5.6)(9.8)(0.866) = 47.53 N
Kf = (47.53)(0.436) = 20.7 N
work done = (20.7)(2.60) = -53.8 J ANS (b)
{neg sign means Kf direction is opposite to motion direction}

The Normal force is the force of the surface of incline perpendicular to block.
As this force is *perpendicular* to the direction of motion, NO WORK is
done by this force. ANS (c)

ie there is NO component of THIS force in the direction of motion.
or
another way of saying the same thing is:
Work = Force x Displacement x Cos Θ => but Cos Θ = 0, because Θ = 90°