Question

4) Use work-energy and conservation formulas to solve: a 500N block slides 2m down a 30° ramp with coefficient of kinetic friction 0.2; show all the math steps to derive the algebraic formulas and calculate; a) Work done by the x-component, and the y-component, of the weight force, and the normal force: b) Work done by the net force c) The final speed of block using the work energy theorem

Answer 1

0 0

Let x component be along the incline and y component be perpendicular to the incline

Along x direction

Work done by weight = mgsin(30°) ×2

= 250 × 2 = 500J

Work done by normal = 0J

Along y direction

Work done by weight = mgcos30° × 2 × cos 90

= 0

Work done by normal = N × 2 × cos90°

=0

The work is zero as both these forces are perpendicular to the displacement.

B) Net force = mgsin(30°) - f

= mgsin(30°) - umgcos30 ° ( because f = uN and N is mgcos30°)

= 500/2 -500 × 0.2 × √3/2 = 163.39 N

Hence work done will be = 163.39 × 2 = 326.8 J

C) Work energy theorem states that the change in Kinetic Energy = Work done by all the forces

1/2 m v ^2 = Work done by all forces = 326.8

v^2 = 326.8 × 2/51.02 = 12.51

v = 3.58 m/s