Q. A body of mass $1\ kg$ is thrown upwards with a velocity $20\ m/s$. It momentarily comes to rest after attaining a height of $18\ m$. How much energy is lost due to air friction $(g = 10\ m/s^{2})$
A) $20\ J$
B) $30\ J$
C) $40\ J$
D) $10\ J$
Explanation
(a) Initial energy of body $= \frac{1}{2}mv^{2} = \frac{1}{2} \times 1 \times (20)^{2} = 200\ J$
A part of this energy consumes in doing work against gravitational force and remaining part consumes in doing work against air friction
i.e., $W_{T} = W_{grav.} + W_{air\ friction}$
$\Rightarrow 200 = 1 \times 10 \times 18 + W_{air} \Rightarrow W_{air} = 20\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.