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Asked by jineshrajbhoi132 | 19 Dec, 2022, 08:44: PM
Expert Answer
Figure shows the forces acting on particle that slides along inclined surface .
Weight ( m g ) is resolved into ( mgsin30 ) along the surface and ( mgcos30) normal to the surface.
Hence Normal reacion force N = mg cos30 and friction force f = μ N = μ mg cos30
where μ is kinetic friction coefficient
Energy loss due to friction in the path PQ = friction force × distance
Energy loss due to friction in the path PQ = μ mg cos30 × (2 / sin30) = ( 2√3 ) μ mg
Energy loss due to friction in the path QR = ( μ mg ) × QR
If energy loss due to friction in both the paths are same , then
( μ mg ) × QR = ( 2√3 ) μ mg
Hence QR = 2√3 m ≈ 3.5 m
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Kinetic energy of particle at Q = ( m g × 2 ) - ( 2√3 μ mg ) = ( 2 m g ) [ 1 - √3 μ ]
Above kinetic energy equals energy loss in the path QR because particle comes to rest at R
( 2 m g ) [ 1 - √3 μ ] = ( μ mg ) × 2√3
we get μ = 0.239 from above expression
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