Explanation

(a) The various forces acting on the block are as shown
As the truck moves in forward direction with acceleration $2\ {m/s}^{2}$, the box experiences a force $F$ in backward direction, [image:5_image272.emf]
$F = ma = 40 \times 2 =$ 80 N
in backward direction.
Its motion will be opposed by force of friction
$f = \mu\ N = \mu\ mg = 0.15 \times 40 \times 10 = 60$ N
The acceleration of the box relative to the truck toward the rear end is
$a = \frac{F - f}{m} = \frac{80 - 60}{40} = 0.5\ m/s^{2}$
If $t$ be the time taken by the box to fall off the truck
$s = ut + \frac{1}{2}{at}^{2}$
$5 = 0 + \frac{1}{2} \times 0.5 \times t^{2}$
$t = \sqrt{20}\ s$
During this time distance covered by truck
$x = 0 \times t + \frac{1}{2} \times 2 \times {(\sqrt{20)}}^{2} = 20\ m$