the upper half of an inclined plane with inclination (theta) is perfectly smooth while the lower half is rough a body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by
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For upper half which is smooth, the acceleration is g×sin(a).
Since the object starts from rest the final velocity when it reaches the end of upper half
is √( g×l×sin(a) ).
For the lower part the acceleration is ( g×sin(a) - μ×g×cos(a) ) due to friction
Since the object comes to rest when it reaches the bottom of inclined plane
we use the relation " u^2 + 2αs = 0 ", where α is acceleration.
Hence ( g×l×sin(a) ) = -2 ( g×sin(a) - μ×g×cos(a) ) (l/2)
Solving for μ we will get μ = 2 × tan(a)
Since the object starts from rest the final velocity when it reaches the end of upper half
is √( g×l×sin(a) ).
For the lower part the acceleration is ( g×sin(a) - μ×g×cos(a) ) due to friction
Since the object comes to rest when it reaches the bottom of inclined plane
we use the relation " u^2 + 2αs = 0 ", where α is acceleration.
Hence ( g×l×sin(a) ) = -2 ( g×sin(a) - μ×g×cos(a) ) (l/2)
Solving for μ we will get μ = 2 × tan(a)
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jot5797:
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