KCET · Physics · Laws of Motion
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 \( \left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right) \)
- A \( 1 \mathrm{~kg} \)
- B \( 2 \mathrm{~kg} \)
- C \( 3 \mathrm{Kg} \)
- D \( 4 \mathrm{Kg} \)
Answer & Solution
Correct Answer
(B) \( 2 \mathrm{~kg} \)
Step-by-step Solution
Detailed explanation
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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