Question

A trolley, having an open rear end, has a small box at mass m, put on it at a distance l from that open rear end. The coefficient of friction between the box and the trolley is μ. If the trolley (initially at rest on a horizontal floor), were to be given a constant acceleration a, the distance, it would travel before the box falls off it, would equal

(1) \(l\frac{(a+μg)}{a}\)

(2) \(l\frac{a}{(a+μg)}\)

(3) \(l\frac{a}{(a-μg)}\)

(4) \(l\frac{(a-μg)}{a}\)

Answer 1

 (3) \(l\frac{a}{(a-μg)}\)

When trolley accelerates from left to right, with an acceleration a, the box has a force (= ma), acting on it from right to left. The free body diagram, showing the forces acting on the box, is as shown. The net force, on the box, directed from right to left, is (ma – μN) = (ma – μ mg) = m(a – μg).

Hence the net acceleratino of the box, directed from right to left, is \(\frac{m(a-μg)}{m}i.e.\) (a-μg).

The box needs to move a distance l (from right to left) before it falls off. If it takes a time t, for this purpose, we would have

The distance, s, moved by the trolley, in this time, is