Question

If the coefficient of friction of a plane inclined at 45° is0.5 then acceleration of a body sliding freely on it is (g=9.8m/s×s

Answer 1

FBD of body,
mgsin∅ - f = ma
mgsin∅ - μmgcos∅ = ma [taking (m) common]
gsin∅ - μgcos∅ = a
9.8xsin45 - (0.5)(9.8)cos45 = a
(9.8)(1/√2) - (0.5)(9.8)(1/√2) = a
(9.8)(1√2)[1-0.5] = a
(9.8/√2)[0.5] = a
(9.8)(0.5)/√2 = a
(4.9)/√2 = a
a = 4.9/√2 m/s²

Answer 2

Answer:

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