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A box when dropped from a certain height reaches the ground with a speed $v$. When it skides from rest from the same height down a rough inclined plane inclined at in angle $45^{\circ}$ to the horizontal, it reaches the ground with a speed $v / 3 .$ The coefficient of sliding friction between the box and the plane is (acceleration due to gravity is $10 \mathrm{~ms}^{-2}$ )
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2985 Upvotes
Verified Answer
The correct answer is:
$\frac{8}{9}$
Case-1
$v=\sqrt{2 g h}$
Case-2
$\Delta U+\Delta k E=w_{f}$
$\begin{array}{l}
-m g h+\frac{1}{2} m\left(\frac{2 g h}{9}\right)=-\mu m g h \\
\mu=\frac{8}{9}
\end{array}$
$v=\sqrt{2 g h}$
Case-2
$\Delta U+\Delta k E=w_{f}$
$\begin{array}{l}
-m g h+\frac{1}{2} m\left(\frac{2 g h}{9}\right)=-\mu m g h \\
\mu=\frac{8}{9}
\end{array}$
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