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The rear side of a truck is open and a b...

The rear side of a truck is open and a box of 40 kg mass is placed 5 m from the openend 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 `2 ms^(-2)` At what distance from the starting point does the box fall off the truck ? (Ignore the size of the box )
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Text Solution

Verified by Experts

The correct Answer is:
`18.57`
Mass of box, `m=40kg`
Acceleration of truck, `a=2ms^(-2)`
Distance of the box from the rear end, `s=5m`
Coefficient of friction,
`mu=0.15`
As the box is in an accelerated frame, it experiences a backward force,
F=ma
Motion of the box in opposed by the frictional force,
`f=muR=mu mg`
`:.` Net force on the box in the backward direction is
`F.=F-f=ma-mu mg=m(a-mu g)`
`=40(2-0.15xx9.8)=21.2N`
Acceleration produced in the box in the backward direction,
`a.=(F.)/(m)=(21.2)/(40)=0.53ms^(-2)`
If the box takes time t to fall off the truck , then
`s=ut+(1)/(2)a.t^(2)` or `5=0xxt+(1)/(2)xx0.53xxt^(2)`
or `t^(2)=(5xx2)/(0.53)=(10)/(0.53)`
This distance covered by the truck accelerating at `2ms^(-2)` during this time is
`s.=(1)/(2)at^(2)=(1)/(2)xx2xx(10)/(0.53)=18.57m`.
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