Question

a block of mass 10 kg is initially at rest when it slides down a frictionless incline whose height is 10 m and is pitched at an angle of 30 degrees. At the bottom of the incline the mass encounters a horizontal surface that has a coefficient of kinetic friction of 0.4 with the mass. How far from the bottom of the incline will the mass come to a stop?

Answer 1

velocity at bottom of the incline

v^2 = 2 g h

v^2 = 2* 9.8* 10

v = 14 m/s

using work-energy theroem

u mg d = 0.5 m v^2

0.4* 9.8* d = 0.5* 14^2

d = 25 m

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