Q. The velocity of a body of mass 20 kg decreases from $20\ {ms}^{- 1}to\ 5\ {ms}^{- 1}$ in a distance of 100 m. Force on the body is
A) $- 27.5\ N$
B) $- 47.5\ N$
C) $- 37.5\ N$
D) $- 67.5\ N$
Explanation
(c) Using Newton’s equation of motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{2} = u^{2} + 2as$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ (5)^{2} = (20)^{2} + 2(a)100$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ a = - \frac{400 - 25}{200} = - \frac{375}{200}{ms}^{- 2}$
Therefore, fore $F = mass\ m\ \times acceleration\ a$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 20 \times \left( - \frac{375}{200} \right) = - 37.5\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.