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

Explanation

When the truck is at rest, the box is at rest. When the truck accelerates, the box likes to be at rest due to inertia. For this it accelerates backwards. But force of friction opposes. So it acts forward i.e., in the direction of motion of truck.
Maximum value of static friction `=mu_(s)mg`
`m=40kg`, `mu_(s)=0.15`, `a=2ms^(-2)`
Net force on box `=ma-mu_(s)mg`
`=40xx2-0.15xx40xx9.8`
`=21.2N` backwards
Backward acceleration of box `=("Net force")/("Mass")=(21.2)/(40)=0.53ms^(-2)`
Now consider the motion of box
First let us find the time taken to travel 5m.
`u=0`, `a=0.53ms^(-2)`, `s=5m`, `t=?`
`s=ut+(1)/(2)at^(2)`, `s=0+(1)/(2)xx0.53xxt^(2)`
`t^(2)=(10)/(0.53)`, `t=sqrt((10)/(0.53))=4.34s`
Thus the box will fall from the back after `=4.34s`
During this time the distance travelled by the truck `S_(t)=?`
For truck `u=0`, `a=2ms^(-2)`, `t=4.34s`
`s_(t)=ut+(1)/(2)at^(2)=0+(1)/(2)xx2xx4.34^(2)=18.86m`