Q. A body is coming with a velocity of $72\ {kmh}^{- 1}$ on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is $10\ {ms}^{- 2}$, find the minimum distance it can be stopped
A) 400 m
B) 40 m
C) 0.40 m
D) 4 m
Explanation
(b) Velocity $u = 72\ {kmh}^{- 1}\ = 20\ {ms}^{- 1}$
$a = \ \mu\ g = 0.5 \times 10\ {ms}^{- 2}$
$From\ \ \ \ \ v^{2} = u^{2} - 2as$
$\therefore\ \ (0)^{2} = (20)^{2} - 2 \times 0.5 \times 10 \times s$
$\therefore s = \frac{20 \times 20}{2 \times 0.5 \times 10}\ or\ s = 40m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.