Question

The rear side of a truck is open and a box of 40kg is placed 5m away from the open and as shown in figure. The coefficient of friction between the box and the surface below it is 0.15. on a straight road, the truck starts and accelerates with $\mathop {2m/s}\nolimits^2 $. At what distance from the starting point does the box fall off the truck? (ignore the size of the box).

Answer 1

Hint: in order to solve this type of question, we will analyse the given equation 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 solution:
Given
Mass of the box, m= 40kg
Coefficient of friction, $\mu = 0.15$
Acceleration of the truck, $\mathop {a = 2m/s}\nolimits^2 $
Distance between the rear side of truck and the box = 5m
Force on the truck
$F = ma$
$F = 40 \times 2 = 80N$
Frictional force on the block
Frictional force is given by, $f = \mu mg$
$f = 0.15 \times 40 \times 10 = 60N$
Therefore, the net force acting on the block is given by,
$\mathop F\nolimits_{net} = F - f$
$\mathop F\nolimits_{net} = 80 - 60 = 20N$
The direction of force is in the backward direction.
Now the backward acceleration,
$\mathop a\nolimits_b = \dfrac{{\mathop F\nolimits_{net} }}{m} = \dfrac{{20}}{{40}} = \dfrac{1}{2} = \mathop {0.5m/s}\nolimits^2 $
Now using the equation of motion we can write,
$S = ut + \dfrac{1}{2}\mathop {at}\nolimits^2 $
$5 = 0 + \dfrac{1}{2} \times \mathop {0.5t}\nolimits^2 $
$ \Rightarrow t = \sqrt {20} s$
The time required to fall from the truck
$S = ut + \dfrac{1}{2}\mathop {at}\nolimits^2 $
$\mathop S\nolimits_t = 0 + \dfrac{1}{2} \times 2 \times (\mathop {\sqrt {20} }\nolimits^2 ) = 20m$

Note: In order to solve this type of question, you need to have a concept of kinematic equation 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 the truck and ultimately it will fall after some time which is calculated above.