Q125 / 160 · AP EAMCET 2023 (15 May Shift 2) · Physics
A block of mass $5 \mathrm{~kg}$ moving on a rough surface with a velocity of $4 \mathrm{~ms}^{-1}$ is stopped by the friction in 2 seconds. Then the coefficient of friction between the contact surfaces is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
- A$0.4$
- B$0.3$
- C$0.5$
- Correct answerD$0.2$
Solution
Mass of block, $\mathrm{m}=5 \mathrm{~kg}$
velocity $\mathrm{v}=4 \mathrm{~m} / \mathrm{s}$
$\begin{aligned}
& \mathrm{F}=\mu \mathrm{mg} \Rightarrow \mathrm{ma}=\mu \mathrm{mg} \\
& \frac{\mathrm{mv}}{\mathrm{t}}=\mu \mathrm{mg} \Rightarrow \mu=\frac{\mathrm{v}}{\mathrm{gt}}=\frac{4}{2 \times 10}=0.2
\end{aligned}$
Hence, the correct answer is option: 'D'.