Question

The rear side of a truck is open and a box of $40\text{kg}$ mass is placed $5\text{m}$ away from the open end as shown. The coefficient of friction between the box and the surface below it is $0.15$. On a straight road, the truck starts from rest and accelerates with $2{\text{ms}}^{-2}$. 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]

• A. $40\text{m}$
• B. $20\text{m}$
• C. $30\text{m}$
• D. $50\text{m}$

Answer 1

The correct option is B $20\text{m}$

Let the acceleration of the box w.r.t truck be $a_{\text{box}}$
Given, truck starts from rest and accelerates with $a=2m/s^2$
In the reference frame of the truck, FBD of $40\text{kg}$ block is as shown


From the FBD, we have,
$N=40g=400\text{N}$
$\Rightarrow f_l=\mu N=0.15\times 400=60\text{N}$
Since $ma=80\text{N}>f_l$, the box slips.

$\therefore$ Net force acting on the box is
$ma-\mu N=40\times 2-60=20\text{N}$
$\Rightarrow m{a}_{\text{box}}=20$
$\Rightarrow a_{\text{box}}=\frac{20}{40}=\frac{1}{2}{\text{m/s}}^2$

So, time taken by box to fall off the truck is given by
$S=ut+\frac{1}{2}{a}_{\text{box}}{t}^2$
$\Rightarrow 5=0+\frac{1}{2}\times \frac{1}{2}\times t^2$
$\Rightarrow t^2=20s$

Thus, distance moved by the truck in that time is
$S=ut+\frac{1}{2}a{t}^2$
$\Rightarrow S=\frac{1}{2}\times a\times t^2$
$\Rightarrow S=\frac{1}{2}\times 2\times \left(20\right)=20\text{m}$