question_answer

A block of mass 'm' is placed on an inclined plane having coefficient of friction 'm?. The plane is making an angle \[\theta \] with the horizontal. The minimum value of upward force acting along the inclined plane that can just move the block up is

A)  \[mg\,\,\cos \,\theta \]        

B)  \[mmg\,\,\cos \,\theta \]

C)  \[mg\,\,sin\,\theta \]         

D)  \[mmg\,\,sin\,\theta \]

Correct Answer: C

Solution :

Suppose a force F is acting on the block as shown in the figure. Minimum value of the force required to just move the block up is      - \[{{F}_{\min }}=mg\,\sin \,\theta +{{f}_{\min }}\] Where, f = frictional force \[{{f}_{\min }}=0\] \[\Rightarrow \] \[{{F}_{\min }}=mg\,\sin \,\theta \]