Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

The initial velocity of box relative to the belt is 3m/s in backward direction. (opposite to motion of belt) ie 3 m/s. The box experience maximum frictional force $F=\mu R=\mu mg$

Therefore $a=\large\frac{Force}{mass}$

$\qquad\qquad=\large\frac{\mu mg}{m}$

$\qquad\qquad=\mu g$

$\qquad\qquad=0.5 \times 9.7 =4.9 m/s^2$

in forward direction

Let the box come to rest in distance S

$v^2=a^2+2as$

$0=(-3)^2+2 \times 4.9 \times s$

$-9=2 \times 4.9 \times S$

$S=-0.918\;m$

We sign shows the box moves in opposite direction of motion of belt

Hence b is the correct answer

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...