A block of mass `2 kg` is placed on a truck as shown in the figure. The coefficient of friction between the block and surface is `0.5`. The truck starts from the rest and moves with acceleration `a_(0)`. (a) Find the friction force acting on the block if (i) `a_(0) = 2 m//s^(2)` , (ii) `4 m//s^(2)` , (iii) `6 m//s^(2)` and (b) if acceleration of truck `a_(0) = 8 m//s^(2)`. After how much time the blocks falls off the truck and the distance travelled by the truck in this time.

Explanation

(a) `a' = mu g = 0.5 xx 10 = 5 m//s^(2)`
If acceleration of the truck `a_(0) le mu g` , the truck and the block move together.
(i) `a_(0) = 2 m//s^(2)`, acceleration of the block ` = a_(0)`
`f = 2a_(0) = 2 xx 2 = 4 N`
(ii) `a_(0) = 4 m//s^(2)`
`f = 2a_(0) = 2 xx 4 = 8 N`
(iii) `a_(0) = 6 m//s^(2) gt mu g` , the block will move in the backward direction with respect to the truck
`N_(0) = 20 N`
`f_(max) = mu N = 0.5 xx 20 = 10 N`
(b) `a_(0) = 8 m//s^(2) gt mu g` , the block will move in the backward direction w.r.t. the truck
`10 = 2a rArr a = 5m//s^(2)`
Acceleration of the block w.r.t. the truck is in the backward direction.
`a_(1) = a_(0) - a = 8 - 5 = 3 m//s^(2)`
`s = (1)/(2) a_(1) t^(2) rArr 6 = (1)/(2) xx 3 xx t^(2) rArr t = 2 s`
The block will fall off the truck in `2 s`. Distance travelled by the truck in this time.
`s' = (1)/(2) a_(0) t^(2) = (1)/(2) xx 8 xx 2^(2) = 16 m`