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 figure. 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 $2 m s^{- 2}$ . Find the distance (in m) travelled by the truck by the time box fall from the truck.(Ignore the size of the box).

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Solution

Value of frictictional force $f$ on the block is
$f = m g = 0.15 \times 40 \times 10 = 60 N$

Let $a_{2}$ be the acceleration of the block towards left

$m a - f = m a_{2}$

$40 \times 2 - 60 = 40 a_{2}$

$40 a_{2} = 20$

$a_{2} = 0.5 m / s^{2}$

Now, time taken for the box to fall is $t$

$5 = \frac{1}{2} a_{2} t^{2}$

$5 = \frac{1}{2} (0.5) t^{2}$

$t^{2} = 20$

Now, distance travel by truck in this time period $t$ is $x$ .

$x = \frac{1}{2} a t^{2}$

$x = \frac{1}{2} (2) (20) = 20 m$

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Q. The rear side of a truck is open and a box of

40 k g

mass is placed

5 m

away from the open end 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 from rest and accelerates with

2 m s^{- 2}

. Find the distance (in m) travelled by the truck by the time box falls from the truck. (Ignore the size of the box).

The rear side of truck is open and a box of 40 kg mass is placed 5m away from its open end. 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 2m/s $^{2}$ . The distance covered by the truck from the starting point until the box falls off (ignore the size of the box)?