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 = 0.15 and g = 10 m/s². The truck starts from rest with an acceleration of 2 m/s² on a straight road. The box will fall off the truck when it is at a distance from the starting point equal to

Maximum acceleration of the box can be μg or 1.5 m/s² while acceleration of truck is 2 m/s². Therefore, relative acceleration of the box will be a_r = 0.5 m/s² (backward). It will fall off the truck in a time.

$$\begin{gathered} t=\sqrt{\frac{2l}{a_r}} \left(s=\frac{1}{2}a{t}^2\right) \\ =\sqrt{\frac{2\times 4}{0.5}}=4s \end{gathered}$$

Displacement of truck up to this instant is

$s_r=\frac{1}{2}{a}_r{t}^2=\frac{1}{2}\times 2\times {\left(4\right)}^2=16 m$