JEE Class main Answered
Sir pls solve the following.
Asked by rsudipto | 22 Dec, 2018, 20:07: PM
Expert Answer
As seen from figure , net force acting on the block = (mg sin30 - μ mg cos30) Newton
Acceleration = g(sin30 - μ cos30) m/s2
distance S travelled along inclined plane in 2 s, starting from rest, S = (1/2)×a×t2 = (1/2)×g(sin30 - μ cos30)×2×2 = 8
after simplifying above equation, we get 9.8×(1- √3μ) = 8 .....(1)
From (1), we get μ = 0.106
Answered by Thiyagarajan K | 22 Dec, 2018, 22:25: PM
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