Question

A block of mass 2 kg starts moving when the angle of inclination of the inclined plane is ${60}^0.$ if the coefficient of kinetic friction is 0.6, the frictional force is?

• A. 2 N
• B. 6 N
• C. 8 N
• D. 10 N

Answer 1

The correct option is B 6 N

$\begin{array}{c}Contactforceistheresul\tan toffrctinalforceandnormalreaction\\ Thefrictionalforcecanbecalculated=\mu kN\\ whereN=mg\cos \theta \\ and,\mu k=coefficientofkineticfriction\\ Now,\\ giventha:-\\ g=10m/s^2\\ M=2kg\\ \mu k=0.6\\ \theta ={60}^0\\ N=\mu k\times mg\cos \theta \\ \Rightarrow 0.6\times 2\times 10\times \cos {60}^0\therefore \left[\cos {60}^0=\frac{1}{2}\right]\\ \Rightarrow 12\times \frac{1}{2}\\ \Rightarrow 6N\end{array}$

Hence,

option $\left(B\right)$ is correct answer.