A Force 18 N pulls a 40 kg mass on a horizontal floor at 0.3 ms\(^{-2}\). Find the coefficient of friction.
a
0.003
b
0.015
c
0.300
d
0.600
Explanation
Correct Option
bVideo Explanation
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PerfectEmerole
5 months ago
The answer is B.
Ma = F-umg (uR)
40×0.3 = 18-u×40×10
12 =18-400u
12-18 =-400u
-6 =-400u
u = -6/-400 the negative signs cancel each other = 0.015
