A) mg sin \[\theta \]
B) \[\mu mg\cos \theta \]
C) \[\mu mg\cos \theta +mg\sin \theta \]
D) \[\mu mg\cos \theta -mg\sin \theta \]
Correct Answer: D
Solution :
The free body diagram showing the various forces acting on the block are as follows: The frictional force\[({{F}_{k}})\]acts opposite to direction of motion of block, given by \[{{F}_{k}}=\mu R\] where R is reaction of the plane on block. \[R=mg\text{ }cos\theta \] \[\therefore \] \[{{F}_{k}}=\mu mg\,\cos \theta \] Also, downward force is \[F=mg\,\sin \theta \] The resultant of F and F provides the necessary acceleration to the block. \[\therefore \]Resultant force upwards \[=\mu mg\cos \theta -mg\,\sin \theta \]You need to login to perform this action.
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