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 in the figure. The coefficient of friction between the box and the surface below is $0.15$. On the straight road, the truck starts from rest and accelerates with $2m/s^2$. Find the distance (in metres) moved by the truck in the time taken by the box to leave the truck. Ignore the size of the box. (Take $\sqrt{10}=3.15$ and$\sqrt{20}=4.47)$

Answer 1

The maximum force of friction between the box and the surface of the truck $=F_s=\mu R=0.15\times 40\times 10=60N.$
This force of friction between the box and the surface of the truck can produce an acceleration of the box $=\frac{60}{40}=1.5m/s^2.$
The acceleration will be in the direction of motion of the truck.
But the truck is accelerating with $2m/s^2$, hence the relative acceleration between the box and the truck $=(2-1.5)=0.5m/s^2.$

In Truck frame.
Now, relative to truck, the box will start moving with the net acceleration of $0.5m/s^2$ towards the open end of the truck. To fall down, the box has to travel a distance of $5m$. The time taken by the box can be calculated using the relation $s=ut+\left(\frac{1}{2}\right)a{t}^2$, we get
$5=0\times t+\left(\frac{1}{2}\right)\times 0.5\times t^2$
$\Rightarrow t^2=\frac{2\times 5}{0.5}=20\Rightarrow t=4.47$ seconds.
Thus, the box will take 4.47 seconds to fall down.
Distance travelled by the truck would be
$s=0\times 4.47+\left(\frac{1}{2}\right)\times 2\times 20=20m$.