Q36 of 40 Page 109

The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. 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 m s-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Given:

Mass of the block, M = 40 Kg


Co-efficient of friction, � = 0.15


Initial velocity, u = 0ms-1


Acceleration, a = 2ms-2


Distance of Box from the end of truck, s = 5m


Using the Newton’s 3rd law of motion, the force on the Box, F can be written as,


F = M× a


F = 40Kg × 2 ms-2


F = 80 N


Since there is friction between the truck and the body, we can write frictional force as,


fs = �× M× g


fs = 0.15× 40Kg × 10ms-2


fs = 60 N.


Since, force of friction is less than that of exerted by the truck, there exists a net force on Box.


FNet = F-fs


FNet = 80 N – 60 N


FNet = 20 N


This force acts in backward direction.


We can find the acceleration of the Box using Newton’s second Law,


FNet = M× a


a = 20 N/ 40 Kg


a = 0.5 ms-2


The box accelerates in backward direction at the rate of 0.5 ms -2.


Since, box is 5 m away from end, i.e. 5 m away from falling down, so we can use the second equation of motion to find the time of fall.


Second equation of motion is given as,


s = ut + 1/2 at2 …1


Where,


s = distance travelled


u = initial velocity


t = time


a = acceleration


Since box was at rest initially so u = 0ms-1


Equation 1 can be re-written as,


S = 1/2 at2


→ t = (2s/a)1/2


→ t = 201/2 s


Now, the distance travelled by truck till the time of fall can also be calculated by equation 1


s = ut + 1/2 at2


s = 0 + 0.5 × 2× (201/2)2


s = 20 m


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