Question

A block of mass $10$ kg is kept on a fixed rough $(\mu =0.8)$ inclined plane of angle of inclination $30$. The frictional force acting on the block is

• A. $50$ N
• B. $50\sqrt{3}$ N
• C. $52$ N
• D. $54$ N

Answer 1

The correct option is A $50$ N

When a block of $10$kg is moving n a rough surface and the angle of inclination is $30$.

the force of static friction is $5$kg .wt

$1$kg.wt means a force that is required to accelerate a mass of $1$kg with an acceleration of approximately $9.8$ $m/s^2$.

here the friction force to stop the block is $mgsin30$$

Since mass is equal to $10$ kg

force required $=10\times sin30$ newton

$\begin{array}{c}=5\times gnewton\\ =5kg.wt\end{array}$

Hence, Option $A$ is correct.