question_answer

A rear side of a truck is open and a box of mass 20 kg is placed on the truck 4 metres away from the open end. \[\mu =0.15\] and\[\text{g }=\text{ 1}0\text{ m}/{{\text{s}}^{\text{2}}}\]. The truck starts from rest with an acceleration of \[\text{2 m}/{{\text{s}}^{\text{2}}}\]on a straight road. The box will fall off the truck when it is at a distance from the starting point equal to

A)  4 metres         

B)         8 metres

C)  16 metres         

D)         32 metres

Correct Answer: C

Solution :

  Frictional force\[=\mu mg\]             \[=0.15\times 20\times 10=30\,\,N\]             \[a=\frac{F}{m}=\frac{10}{20}=0.5\,\,m/{{s}^{2}}\] Now,     \[s=\frac{1}{2}a{{t}^{2}}\] \[\therefore \]      \[4=\frac{1}{2}\times 0.5{{t}^{2}}\] \[\Rightarrow \]   \[t=4\,\,\sec \] Now distance of truck from initial position after             \[t=4\,\,\sec \]             \[s=\frac{1}{2}{{a}_{1}}{{t}^{2}}=\frac{1}{2}\times 2\times 16=16\,\,m\]