__Question 5.36__: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).

__Solution__:Mass of the box,

*m*= 40 kg

*μ*= 0.15

Initial velocity,

*u*= 0

Acceleration,

*a*= 2 m/s

^{2}

Distance of the box from the end of the truck,

*s*' = 5 m

As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by:

*F*=

*ma*

*=*40 × 2 = 80 N

As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction

*f*, acting between the box and the floor of the truck. This force is given by:

*f*=

*μ*

*m*g

= 0.15 × 40 × 10 = 60 N

∴Net force acting on the block:

*F*

_{net}= 80 – 60 = 20 N backward

The backward acceleration produced in the box is given by:

*a*

_{back}= F

_{net}/ m = 20 / 40 = 0.5 ms

^{-2}

Using the second equation of motion, time

*t*can be calculated as:

s' = ut + (1/2)a

_{back}t

^{2}

5 = 0 + (1/2) X 0.5 X t

^{2}

^{}

∴ t = √

*20*s

Hence, the box will fall from the truck after √

*20*s from start.

The distance

*s*, travelled by the truck in √

*20*s is given by the relation:

s = ut + (1/2)at

^{2}

= 0 + (1/2) X 2 X (√

*20)*

^{2}

^{}

= 20 m

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