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A body of mass $2 \mathrm{~kg}$ is on an inclined plane of inclination $30^{\circ}$ and coefficient of friction is $\left(\frac{1}{\sqrt{3}}\right)$. The minimum force required to move the body up the inclined plane is
$\left(\right.$ Acceleration due to gravity $\left.=10 \mathrm{~ms}^{-2}\right)$
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$\left(\right.$ Acceleration due to gravity $\left.=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$20 \mathrm{~N}$
Mass of body, $m=2 \mathrm{~kg}$
For body moving up
$$
\begin{aligned}
& F=m g \sin \theta+f_r \\
& =m g \sin \theta+\mu_k N=m g \sin \theta+\mu_k m g \cos \theta \\
& =2 \times 10 \times \frac{1}{2}+\frac{1}{\sqrt{3}} \times 2 \times 10 \times \frac{\sqrt{3}}{2}=20 N
\end{aligned}
$$
For body moving up
$$
\begin{aligned}
& F=m g \sin \theta+f_r \\
& =m g \sin \theta+\mu_k N=m g \sin \theta+\mu_k m g \cos \theta \\
& =2 \times 10 \times \frac{1}{2}+\frac{1}{\sqrt{3}} \times 2 \times 10 \times \frac{\sqrt{3}}{2}=20 N
\end{aligned}
$$
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