Q. A stone weighing $1\ kg$ and sliding on ice with a velocity of $2\ m/s$ is stopped by friction in $10\ sec$.The force of friction (assuming it to be constant) will be
A) $- 20\ N$
B) $- 0.2\ N$
C) $0.2\ N$
D) $20\ N$
Explanation
(b) $u = 2m/s,v = 0,\ t = 10\ sec$
$\therefore a = \frac{v - u}{t} = \frac{0 - 2}{10} = - \frac{2}{10} = - \frac{1}{5} = - 0.2\ m/s^{2}$
$\therefore$ Friction force $= ma = 1 \times ( - 0.2) = - 0.2\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.