the maximum acceleration of the train in which a box lying on its floor will remain stationary given that the coefficient of friction between the box and the train is 0.15(g=ms^-2? ) is
(a) 15ms^-2, (b) 1.5 ms ^ -2, (C) 2.0ms^ -2, (d) 0.15ms^-2
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Answer:
acceleration a is given by
a = coefficient of friction × g
( Since ma = coefficient of friction × mg)
a = 0.15 × 10 = 1.5m/s^2
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