Explanation

(a) [image:5_image244.png]
Here, Mass of the box, $M = 40\ kg$
Acceleration of the truck, $a = 2\ ms^{- 2}$
Distance of the box from the rear end , $d = 5m$
Coefficient of friction between the box and the surface below it, $\mu = 0.15$
The various forces acting on the block are as shown in the figure
[image:5_image245.emf]
As the truck moves in forward direction with the acceleration $a = 2ms^{- 2}$, the box experiences a force $F$ in the backward direction and it is given by
$f = Ma = (40\ kg) \times \left( 2\ ms^{- 2} \right) = 80\ N$ in backward direction
Under the action of this force, the box will tend to move toward the rear end of the truck . As it does so, its motion will be opposed by the force of friction which acts in the forward direction and it is given by
$f = \mu N = \mu Mg = 0.15 \times 40 \times 10 = 60\ N$
The acceleration of the box relative to the truck toward the rear end is, $a_{1} = \frac{F - f}{M} = \frac{80\ N - 60\ N}{40\ kg} = 0.5\ ms^{- 1}$
Let $t$ be the time taken for the box to fall off the truck
Using, $S = ut + \frac{1}{2}\ at^{2}$, we get , $d = 0 \times t + \frac{1}{2}a_{1}t^{2}\lbrack\because u = 0\rbrack$
$5 = \frac{1}{2} \times 0.5 \times t^{2},\ t = \sqrt{\frac{2 \times 5}{0.5}} = \sqrt{20}\ s$
During this time, the truck covers a distance $x$
Using $S = ut + \frac{1}{2}at^{2}$
We get $x = 0 \times t + \frac{1}{2} \times 2 \times \left( \sqrt{20} \right)^{2}$ $\lbrack\because u = 0\rbrack$
$x = 20\ m$