Question

The rear side of a truck is open and a box of mass 20kg is placed on the truck 4m away from the open end. The coefficient of friction between box and truck surface is 0.15. The truck starts from rest with an acceleration of $2m/{s^2}$ on the straight road. The box will fall of the truck when it is a distance (from starting point) ( take g = $10m/{s^2}$ )
A. 14m
B. 8m
C. 82m
D. 16m

Answer 1

In order to solve this type of question, we will analyze the given question taking the truck as a frame of reference. And since it is a non-inertial frame therefore we will have an extra force on block i.e. pseudo force whose magnitude will be mass (block) x acceleration of truck and direction will be just opposite to direction of acceleration of truck. We will use this concept to solve this question.

Complete step-by-step answer:
Given
Mass of the box, m = 20kg
Coefficient of friction $\mu = 0.15$
Acceleration of the truck $a = 2m/{s^2}$
Distance of between the rear side of truck and the box = 4m
Force on the truck
$F = ma \\\ F = 20 \times 2 \\\ F = 40N \\\$
The frictional force on the block
Frictional force is given by $f = \mu mg$
$f = 0.15 \times 20 \times 10 \\\ f = 30N \\\$
Therefore the net force acting on the block is given by
${F_{net}} = F - f \\\ {F_{net}} = 40 - 30 \\\ {F_{net}} = 10N \\\$
The direction of the force is in a backward direction.
Now the backward acceleration
${a_b} = \dfrac{{{F_{net}}}}{m} = \dfrac{{10}}{{20}} = 0.5m/{s^2}$
The time required by the box to fall from the truck
$S = ut + \dfrac{1}{2}a{t^2} \\\ 4 = 0 + \dfrac{1}{2}0.5{t^2} \\\ {t^2} = 16 \\\ t = 4s \\\$
The distance cover by the truck in 4s is
${S_t} = ut + \dfrac{1}{2}a{t^2} \\\ {S_t} = 0 + \dfrac{1}{2} \times 2 \times {4^2} \\\ {S_t} = 16m \\\$
Hence, the correct option is D.

Note: In order to solve this type of question, you need to have a concept of kinematic equations of motions and newton laws of motion. In the above problem the truck moves forward with some acceleration it exerts some amount of force on block also and that force is force of friction which will bring the block together. But due to insufficient in magnitude it will not have the same acceleration as that of a truck and ultimately it will fall after some time which is calculated above.