Question

The rear side of a truck is open and a box of $40kg$ mass is placed $5m$ away from the open end as shown. The coefficient of friction between the box & the surface below it is $0.15$. On a straight road, the truck starts from rest and accelerates with $2m{s}^{-2}$. At what distance from the starting point does the box fall off the truck (i.e. distance travelled by the truck)? [Ignore the size of the box]

Answer 1

In the reference frame of the truck, FBD of box is


From the FBD, we have
$N=40g=400\text{N}$
$f=\mu N=0.15\times 400=60\text{N}$
Thus, net force $=ma-f=40\times 2-60=20\text{N}$
i.e $m{a}_{box}=20$
$\Rightarrow a_{box}=\frac{20}{40}=0.5m/s^2$

This acceleration of the box is in the reference frame of the truck i.e $a_{box}=a_{rel}$. So, time taken by box to fall down from the truck is given by
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$\Rightarrow 5=0+\frac{1}{2}\times 0.5\times t^2$
$\Rightarrow t^2=20$

Hence, distance moved by the truck is
$S_{truck}=u_{truck}t+\frac{1}{2}\times a_{truck}\times t^2$
$\Rightarrow S_{truck}=0+\frac{1}{2}\times 2\times \left(20\right)=20\text{m}$