Question

A block rests on a rough inclined plane making an angle of ${30}^0$ 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 (in kg) is (take, $\mathrm{g}=10{\mathrm{ms}}^{-2}$)

• A. $2.0$
• B. $4$
• C. $1.6$
• D. $2.5$

Answer 1

The correct option is A

$2.0$

Step 1: Given data and drawing the free body diagram

Coefficient of static friction ${\mathrm{\mu }}_{\mathrm{s}}=0.8$

Frictional force on the block $\mathrm{f}=10\mathrm{N}$

Angle with the horizontal plane $\mathrm{\theta }={30}^0$

And $\mathrm{g}=10{\mathrm{ms}}^{-2}$

The free body diagram of the block is given below:

Step 2: Calculating the mass of the block

Now, the frictional force on the block at rest is equal to:

$$\begin{gathered} \mathrm{f}=\mathrm{mgsin}\left(\mathrm{\theta }\right) \\ \Rightarrow 10\mathrm{N}=\mathrm{m}\times 10{\mathrm{ms}}^{-2}\times \sin {\left(30\right)}^0 \\ \Rightarrow \mathrm{m}=\frac{1}{\sin {\left(30\right)}^0}\mathrm{kg}=\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \Rightarrow \mathrm{m}=2\mathrm{Kg} \end{gathered}$$

Therefore mass of the block is equal to $2\mathrm{Kg}$

Hence, option(A) is the correct answer.