First slide

Friction on horizontal surface

Question

The rear side of a truck is open and a box of mass 20 kg is placed on the truck 4 m away from the open
end ( $μ$ = 015 and g = 10 $m s^{- 2}$ ). The truck starts from rest with an acceleration of 2 $m s^{- 2}$ on a straight
road. The box will fall off the truck when it is at a distance from the starting point equal to

Moderate

A

4 m
B

8 m
C

16 m
D

32 m

Solution

Maximwn acceleration of the box can be /!g or 1.5 $m s^{- 2}$ , while acceleration of truck is 2 $m s^{- 2}$ . Therefore, relative acceleration of the box will be a_r = 0.5 $m s^{- 2}$ (backward). It will fall off the truck in a time,
$t = \sqrt{\frac{2 l}{a_{r}}} = \sqrt{\frac{2 \times 4}{0.5}} = 4 s$ $(∵ s = \frac{1}{2} a t^{2})$
Displacement of truck upto this instant,
$s_{r} = \frac{1}{2} a_{r} t^{2} = \frac{1}{2} \times 0.5 \times (4)^{2} = 4 m$

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