Question

Starting from rest, a 4.60-kg block slides 2.00 m down a rough 30.0° incline. The coefficient of kinetic friction between the block and the incline is

μ_k = 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

(d) Qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height?

Answer 1

m = 4.6 kg

S = 2 m

angle = 30 degree

Force down the incline due to weight = mg sin 30 = 4.6*9.81*sin 30 = 22.563 N

Normal force to incline = mg cos 30 = 4.6*9.81*cos 30 = 39.08 N

Frictional force= Nμ = 39.08*0.436 = 17.04 N

a) So work done by gravity = 22.563*2 = 45.126 J

b) work done by friction = - 17.04*2 = - 34.08 J (negative work since direction of motion is opposite to frictional force direction)

c) work done by normal force = 0 (since normal is perpendicular to direction of motion)

d) If shorter-steeper ramp is used to span the same vertical height, normal force is reduced, as a result frictional force will reduce.

Work done by force of gravity will remain same for same vertical height

work done by frictional force will reduce, as both the frictional force as well as distance traveled reduces

Work done by normal force will remain 0 as normal is perpendicular to motion in this case too.