9th Class
Science
Force and laws of motion
Question Bank
Force and Laws of Motion
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 \[t=4\,\,\sec \] \[s=\frac{1}{2}{{a}_{1}}{{t}^{2}}=\frac{1}{2}\times 2\times 16=16\,\,m\]