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A body is moving along a rough horizontal surface with an initial velocity of $10 \mathrm{~ms}^{-1}$. If the body comes to rest after travelling a distance of 12 m , then the coefficient of sliding friction will be
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The correct answer is:
0.4
Given, $u=10 \mathrm{~ms}^{-1}, s=12 \mathrm{~m}, v=0$
By third equation of motion,
$\begin{aligned} v^2 & =u^2-2 a s \\ 0 & =(10)^2-2 \times a \times 12 \\ 24 a & =100 \\ a & =4.17 \mathrm{~ms}^{-2}\end{aligned}$
Coefficient of sliding friction is given by
$\mu=\frac{a}{g}=\frac{4.17}{10}=0.41$
By third equation of motion,
$\begin{aligned} v^2 & =u^2-2 a s \\ 0 & =(10)^2-2 \times a \times 12 \\ 24 a & =100 \\ a & =4.17 \mathrm{~ms}^{-2}\end{aligned}$
Coefficient of sliding friction is given by
$\mu=\frac{a}{g}=\frac{4.17}{10}=0.41$
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