Q. A ship weighing $0.3 \times 10^{8}$ kg-wt is pulled by a force of $0.5 \times 10^{5}$ N through a distance of 3 m. If the ship were originally at rest and water-resistance is negligibly small, then the ship will acquire a speed of
A) $0.1\ ms^{- 1}$
B) $1\ ms^{- 1}$
C) $1.5\ ms^{- 1}$
D) $12\ ms^{- 1}$
Explanation
(a) $m = 0.3 \times 10^{8}$ kg, $F = 0.5 \times 10^{5}$N, $s = 3m,\ v = ?$
Work done$= \ F \times s$
This work becomes the kinetic energy of the ship
$\therefore\ \frac{1}{2}mv^{2} = F \times s$
or $v^{2} = \frac{2Fs}{m} = \frac{2 \times 0.5 \times 10^{5} \times 3}{0.3 \times 10^{8}}$ or $v = 0.1ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.