Question

A block of mass $2\mathrm{Kg}$ is at rest on a floor. The coefficient of static friction between block and the floor is $0.54$. A horizontal force of $2.8\mathrm{N}$ is applied to the block. What should be the frictional force between the block and the floor? (take, $\mathrm{g}=10{\mathrm{ms}}^{-2}$)

• A. $8.8\mathrm{N}$
• B. $5.8\mathrm{N}$
• C. $2.8\mathrm{N}$
• D. $10.8\mathrm{N}$

Answer 1

The correct option is C

$2.8\mathrm{N}$

Step 1: Given data

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

Coefficient of static friction, $\mathrm{\mu }=0.54$

Horizontal force applied on the block, $\mathrm{F}=2.8\mathrm{N}$

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

Step 2: Calculating the frictional force between the block and the floor

The above diagram shows the arrangement of block and the force applied on it.

Now, the force of friction is given by, $\mathrm{f}=\mathrm{\mu N}$ and normal force is given by, $\mathrm{N}=\mathrm{mg}$. Therefore we have,

$\mathrm{f}=\mathrm{\mu N}=\mathrm{\mu mg}=0.54\times 2\mathrm{kg}\times 10\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right.=10.8\mathrm{N}$

Here, we see that the frictional force is greater than applied force on the block. So here, the frictional force will adjust itself because frictional force can never be greater than applied force.

Therefore, the frictional force will be equal to $2.8\mathrm{N}$.

Hence, option(C) is correct answer.