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A body is travelling with $10 \mathrm{~ms}^{-1}$ on a rough horizontal surface. It's velocity after $2 \mathrm{~s}$ is $4 \mathrm{~ms}^{-1}$. The coefficient of kinetic friction between the block and the plane 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:
$0.3$
Initial speed of body on the horizontal rough surfaces, $u=10 \mathrm{~ms}^{-1}$
Final velocity, $v=4 \mathrm{~ms}^{-1}$ and $t=2 \mathrm{~s}$.
If $a$ be the retardation due to friction, then by using equation,
$v=u-a t$
$\Rightarrow \quad 4=10-a \times 2$
$\Rightarrow \quad 2 a=6 \Rightarrow a=3 \mathrm{~ms}^{-2}$
According to given situation,
friction force $=m a$
$\begin{aligned} & \Rightarrow \quad \mu_k m g=m a \\ & \Rightarrow \quad \mu_k=\frac{a}{g}=\frac{3}{10}=0.3\end{aligned}$
Final velocity, $v=4 \mathrm{~ms}^{-1}$ and $t=2 \mathrm{~s}$.
If $a$ be the retardation due to friction, then by using equation,
$v=u-a t$
$\Rightarrow \quad 4=10-a \times 2$
$\Rightarrow \quad 2 a=6 \Rightarrow a=3 \mathrm{~ms}^{-2}$
According to given situation,
friction force $=m a$
$\begin{aligned} & \Rightarrow \quad \mu_k m g=m a \\ & \Rightarrow \quad \mu_k=\frac{a}{g}=\frac{3}{10}=0.3\end{aligned}$
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