Question

Maximum acceleration of the train for which a 50 kg box lying on its floor will remain stationary (given:coefficient of static friction between the box and the train's floor is 0.3 and g=10m/s^2)

Answer 1

your answer should be 3m/s^2

If block want to remain stationary forward force must be equal to backward force.

Forward force = Backward(frictional) force

ma. = umg. (u=coeffcient of friction)
50×a. = 0.3×50×10
50a. = 150

a = 150/50
a = 3m/s^2

Answer 2

Explanation:

Hope it will help u.

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