Question

A block of mass $15kg$ is placed at a distance of $10m$ from the rear end of a long trolley as shown in figure. The coefficient of friction between the block and the surface below is $\mathrm{0.4.}$ The trolley initally at rest is given a uniform acceleration of $5m/s^2$ . At what distance (in metres) from the starting point will the block fall-off the trolley? (Take $g=10m/s^2$ )

Answer 1

Answer: 50

Acceleration of the block = acceleration of the trolley $=a=5m/s^2$
The force acting on the block is, $F=ma=15\times 5=75N$ in the backward direction. The force of limiting friction $f$ is given by, $f=\mu mg=0.4\times 15\times 10=60N$ in forward
direction. Hence, the net force on the block towards the right i.e., towards the rear end of the trolley is, $F^{{}^{\prime }}=F-f=75-60=15N$
Acceleration due to this force is, $a^{{}^{\prime }}=\frac{F^{{}^{\prime }}}{m}=\frac{15}{15}=1m/s^2$
Let $t$ be the time taken for the block to fall from the rear end of the trolley travelling a distance $s=10m$ $s=ut+\frac{1}{2}a{t}^2$
$10=0+\frac{1}{2}\times 1\times t^2$
$\therefore t=\sqrt{20}s$
The distance covered by the trolley in time $t=\sqrt{20}s$ is given by $s^{{}^{\prime }}-ut+\frac{1}{2}a{t}^2-0+\frac{1}{2}\times 5\times 20$
$\therefore s^{{}^{\prime }}=50m$