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Asked by sudheerkapoor67 | 25 Jan, 2018, 04:35: PM
answered-by-expert Expert Answer
 
for upper part of inclined plane:-
 
if v is the final velocity when the object reaches vertical height h/2, then v2 = 2gsinΦ (h/2) = ghsinΦ
 
for lower part of inclined plane:-
 
the motion is retarded motion due to friction, retardation is μgcosΦ-gsinΦ.
 
since final velocity is zero, we get
 
0 = ghsinΦ - 2 (μgcosΦ-gsinΦ)(h/2) = ghsinΦ -  gh(μcosΦ-sinΦ)
 
By simplifying the above relation, we get μ = 2 tanΦ
Answered by Thiyagarajan K | 25 Jan, 2018, 11:04: PM

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