a block is pushed up a rough inclined plane. It stops after ascending a few meters and then reverses its direction and returns back to the point where it started. If angle of inclination is 37 degrees and the time to climb up is half of the time to return back , then the coefficient of friction is?

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Solution

$F o r u p w a r d m o t i o n t h r o u g h t h e i n c l i n e d p l a n e : f o r c e i s g i v e n b y : F = m g \sin θ + f r i c t i o n a l f o r c e F = m g \sin θ + μ m g \cos θ a c c e l e r a t i o n a = g \sin θ + μ g \cos θ w h e r e θ i s t h e a n g l e o f i n c l i n a t i o n . d i s \tan c e t r a v e l l e d = \frac{1}{2} a t^{2} w h e r e t i s t h e t i m e t a k e n t o r e a c h h e i g h e s t p o i n t . w h e n c o m i n g d o w n a' = g \sin θ - μ g \cos θ d i s \tan c e t r a v e l l e d = \frac{1}{2} a' t'^{2} t = 2 t' \frac{1}{2} a t^{2} = \frac{1}{2} a' t'^{2} 4 a t'^{2} = a' t'^{2} 4 a' = a' 4 g \sin θ + 4 μ g \cos θ = g \sin θ - μ g \cos θ 5 μ g \cos θ = - 3 g \sin θ μ = - \frac{3}{5} \tan (37) w h e r e μ i s t h e c o f f i c i e n t o f f r i c t i o n .$

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