Question

A block of mass m starts slipping from top of inclined plane at B and comes to rest when reaches to lowest point of the inclined plane, if BC = 2 AC & friction coefficient of part AC is µ=k tan then , find the value of k .

Answer 1

A block of mass m starts slipping from the top of inclined plane at B and comes to rest when reaches to the lowest point of the inclined plane.

here BC = 2AC and coefficient of friction, μ = k tanθ

To find : The value of k

solution : see diagram,

if we assume AC = l, BC = 2l then AB = AC + BC = 3l

now, BT = 3lsinθ [ see figure ]

from work energy theorem,

workdone = change in kinetic energy

⇒work done by gravity - workdone by friction = 0

⇒mg × 3lsinθ - μmgcosθ × l = 0

⇒3sinθ = μcosθ

⇒3tanθ = μ

on comparing we get, k = 3

Therefore the value of k = 3