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If the coeffiecient of friction of a plane inclined at 45 degree is 0.5 , then acceleration of a body sliding freely on it is ( g = 9.8 m/ s^2)
a) 4.9 m/s^2
b) 9.8 m/s^2
c) 9.8/√2 m/s^2
d) 9.8/2√2 m/s^2
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Refer to the attachment...
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Answer:
(MgSin45 -uMgCos45)/M = 9.8/2√2
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