question_answer

A block of mass 2 kg is lying on an inclined plane, inclined to the horizontal at 30°. If the coefficient of friction between the block and the plane is 0.7, then magnitude of frictional force acting on the block will be :

A)  11.9 N                                 

B)  1.19 N

C)  0.19 N                                 

D)  1109 N

Correct Answer: A

Solution :

                The various forces acting on the block are as shown Frictional force\[=\mu R\]                ... (i) where\[\mu \]is coefficient of friction and R the normal reaction of the surface on the block. Also,       \[=R=mg\text{ }cos30{}^\circ \]                            ... (ii) From Eqs. (i) and (ii), we get \[F=\mu \text{ }mg\text{ }cos\text{ }30{}^\circ \] Given, \[\mu =0.7,\text{ }m=2\text{ }kg,\text{ }g=9.8\text{ }m/{{s}^{2}},\]                 \[\cos {{30}^{o}}=\frac{\sqrt{3}}{2}\] \[\therefore \]  \[F=0.7\times 2\times 9.8\times \frac{\sqrt{3}}{2}=11.9N\]