Question

A force of $100N$ is applied on a block of mass $3kg$ as shown in the figure. THe coefficient of the friction between the surface and the block is $\mu =\frac{1}{\sqrt{3}}$. The frictional force acting on the block is

• A. $10m/s^2\text{upwards}$
• B. $10m/s^2\text{downward}$
• C. $\text{zero}$
• D. $\text{none of these}$

Answer 1

The correct option is C $\text{zero}$

Draw free body diagram of the given problem.

Find the acceleration of the block.
It is clear from the diagram that,
$N=100\cos {30}^{\circ }=100.\frac{\sqrt{3}}{2}=50\sqrt{3}N$
Net driving force $=(100\times \frac{1}{2}-30)$
= $20N\text{(upward)}$
$f_{lim}=\frac{1}{\sqrt{3}}\left(50\sqrt{3}\right)=50N$
As the component of applied force is less than the limiting friction, hence block will remain stationary.
AS friction is static nature hence acceleration of the block will be zero.
Final Answer:$\left(A\right)$