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Question:
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A block rests on a rough inclined plane making an angle of \( 30^{\circ} \) with the horizontal. The
coefficient of static friction between the block and the plane is \( 0.8 \). If the frictional force on the
block is \( 10 \mathrm{~N} \), the mass of the block is \( \left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right) \)
Options:
coefficient of static friction between the block and the plane is \( 0.8 \). If the frictional force on the
block is \( 10 \mathrm{~N} \), the mass of the block is \( \left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right) \)
Solution:
2635 Upvotes
Verified Answer
The correct answer is:
\( 2 \mathrm{~kg} \)
Given, coefficient of static friction, \( \mu=0.8 \); frictional force, \( f=10 \mathrm{~N} \)
Now, \( f=m g \sin \theta \)
\[
\begin{array}{l}
\Rightarrow 10=m \times 10 \times \sin 30^{\circ} \\
\Rightarrow 10=m \times 10 \times \frac{1}{2} \\
\Rightarrow m=\frac{10}{5}=2
\end{array}
\]
Therefore, mass of the block is \( 2 \mathrm{~kg} \)
Now, \( f=m g \sin \theta \)
\[
\begin{array}{l}
\Rightarrow 10=m \times 10 \times \sin 30^{\circ} \\
\Rightarrow 10=m \times 10 \times \frac{1}{2} \\
\Rightarrow m=\frac{10}{5}=2
\end{array}
\]
Therefore, mass of the block is \( 2 \mathrm{~kg} \)
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