question_answer

A block of mass m is placed on an inclined plane having coefficient of friction W. The plane is making an angle 0 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 \[{{F}_{\min }}=mgsin\theta +{{f}_{\min }}\] where, \[f=\]frictional force \[{{f}_{\min }}=0\] \[\Rightarrow \] \[{{F}_{\min }}=mg\sin \theta \]