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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 $=10 \mathrm{~ms}^{-2}$ )
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(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$0.2$
Mass of block, $\mathrm{m}=5 \mathrm{~kg}$
velocity $\mathrm{v}=4 \mathrm{~m} / \mathrm{s}$
$\begin{aligned}
& \mathrm{F}=\mu \mathrm{mg} \Rightarrow \mathrm{ma}=\mu \mathrm{mg} \\
& \frac{\mathrm{mv}}{\mathrm{t}}=\mu \mathrm{mg} \Rightarrow \mu=\frac{\mathrm{v}}{\mathrm{gt}}=\frac{4}{2 \times 10}=0.2
\end{aligned}$
velocity $\mathrm{v}=4 \mathrm{~m} / \mathrm{s}$
$\begin{aligned}
& \mathrm{F}=\mu \mathrm{mg} \Rightarrow \mathrm{ma}=\mu \mathrm{mg} \\
& \frac{\mathrm{mv}}{\mathrm{t}}=\mu \mathrm{mg} \Rightarrow \mu=\frac{\mathrm{v}}{\mathrm{gt}}=\frac{4}{2 \times 10}=0.2
\end{aligned}$
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