Question

A block of mass $2\mathrm{Kg}$ rests on a horizontal surface. If a horizontal force of $5\mathrm{N}$ is applied on the block, the frictional force on it is ( take, ${\mathrm{\mu }}_{\mathrm{k}}=0.4$, ${\mathrm{\mu }}_{\mathrm{s}}=0.5$ )

• A. $5\mathrm{N}$
• B. $10\mathrm{N}$
• C. $8\mathrm{N}$
• D. $\mathrm{Zero}$

Answer 1

The correct option is A

$5\mathrm{N}$

Step 1: Given data

Mass of a block, $\mathrm{m}=2\mathrm{Kg}$

Applied horizontal force, $\mathrm{F}=5\mathrm{N}$

Coefficient of static friction, ${\mathrm{\mu }}_{\mathrm{s}}=0.5$

Coefficient of dynamic friction, ${\mathrm{\mu }}_{\mathrm{k}}=0.4$

Acceleration due to gravity, $\mathrm{g}=10{\mathrm{ms}}^{-2}$

Step 2: Finding the frictional force

The frictional force is calculated using the formula:

$$\begin{gathered} \mathrm{f}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{N} \\ \Rightarrow \mathrm{f}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg}(\because \mathrm{N}=\mathrm{mg}) \\ \Rightarrow \mathrm{f}=0.5\times 2\times 10 \\ \Rightarrow \mathrm{f}=10\mathrm{N} \end{gathered}$$

Here, we observe that the frictional force is greater than the applied force, that is, $\mathrm{f}>\mathrm{F}$, hence the frictional force will adjust its value and will be equal to applied horizontal force, that is,

Frictional force, $\mathrm{f}=5\mathrm{N}$

Hence, option(A) is correct answer.