Question

As shown in the figure, a truck has its rear side open and a box of $40\text{kg}$ mass is placed $5\text{m}$ away from the open end. 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 $2{\text{m/s}}^2$. Find the distance (in metres) travelled by the truck by the time the box falls from the truck. (Ignore the size of the box). [Take $g=10{\text{m/s}}^2$]

• A. $10\text{m}$
• B. $5\text{m}$
• C. $20\text{m}$
• D. $15\text{m}$

Answer 1

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

Given, Mass of the block $=40\text{kg}$
$\mu =0.15$
Acceleration of the truck $=a=2{\text{m/s}}^2$

F.B.D of $40\text{kg}$ block (from the frame of reference of the block):


$ma-f_k=m\times a^{\prime }$
(assume $a^{\prime }$ is the acceleration of the block towards right)
$f_k=\mu mg=0.15\times 40\times 10=60\text{N}$

$\Rightarrow 40\times 2-60=40\times a^{\prime }$
$\Rightarrow a^{\prime }=0.5{\text{m/s}}^2$

Time taken by $40\text{kg}$ block to travel $5\text{m}$ distance is given by
$s=ut+\frac{1}{2}a{t}^2$. Here $u=0$
$\Rightarrow 5=0+\frac{1}{2}\times 0.5\times t^2$
$t=\sqrt{20}\text{sec}$

In time $t=\sqrt{20}\text{sec}$, distance travelled by truck.
$s_1=ut+\frac{1}{2}a{t}^2$
Here $u=0$, initial speed of truck
$s_1=0+\frac{1}{2}\times 2\times (\sqrt{20}{)}^2$
$s_1=20\text{m}$