A body of mass ' $m$ ' dropped from a height ' $h$ ' reaches the ground with a speed of $0.8 \sqrt{ } \mathrm{gh}$. The value of workdone by the air-friction is
Select the correct option:
A
–0.68 mgh
B
mgh
C
) 1.64 mgh
D
0.64 mgh
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Work done = Change in kinetic energy
$$
\begin{aligned}
& \mathrm{W}_{\mathrm{mg}}+\mathrm{W}_{\mathrm{ar} \text {-fiction }}=\frac{1}{2} \mathrm{~m}(0.8 \sqrt{\mathrm{gh}})^2-\frac{1}{2} \mathrm{~m}(0)^2 \\
& \mathrm{~W}_{\text {air-frition }}=\frac{.64}{2} \mathrm{mgh}-\mathrm{mgh}=-0.68 \mathrm{mgh}
\end{aligned}
$$
Option (1)
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
