Question

the rear side of a truck is open and a box of 40 kg is placed 5m away from the open end . The friction between the box and the surface below is 0.15 on a straight horizontal road the truck starts from rest and accelerates with 2 m/s at what distance from the starting point the box fall off the truck ignore the mass of the truck

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Mass}\mathrm{of}\mathrm{box},\mathrm{m}=40\mathrm{kg} \\ \mathrm{Distance}\mathrm{from}\mathrm{open}\mathrm{end},\mathrm{s}=5\mathrm{m} \\ \mathrm{Coefficient}\mathrm{of}\mathrm{friction},\mathrm{\mu }=0.15 \\ \mathrm{Acceleration}\mathrm{of}\mathrm{truck},\mathrm{a}=2\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Frictional}\mathrm{force},\mathrm{f}=\mathrm{\mu mg}=0.15\times 40\times 10=60\mathrm{N} \\ \mathrm{Pseudo}\mathrm{force}\mathrm{on}\mathrm{the}\mathrm{block}\mathrm{will}\mathrm{be}\mathrm{given}\mathrm{by}, \\ \mathrm{F}=\mathrm{ma}=40\times 2=80\mathrm{N} \\ \mathrm{Acceleration}\mathrm{of}\mathrm{block},\mathrm{a}=\frac{\mathrm{F}-\mathrm{f}}{\mathrm{m}}=\frac{80-60}{40}=0.5\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Suppose}\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{is}\mathrm{t} \\ 5=0\times \mathrm{t}+\frac{1}{2}\times 0.5\times {\mathrm{t}}^2 \\ \mathrm{or}\mathrm{t}=\sqrt{20}=4.34\mathrm{s} \\ \mathrm{Distance}\mathrm{travelled}\mathrm{by}\mathrm{truck}\mathrm{will}\mathrm{be}\mathrm{given}\mathrm{by}, \\ \frac{1}{2}\times 2\times 4.{34}^2=18.84\mathrm{m} \end{gathered}$$