question_answer

An inclined plane has an inclination \[\theta \]with horizontal. A body of mass m rests on it. If the coefficient of friction between the body and the plane is a, then the minimum force that needs to be applied parallel to the inclined plane is

A)  \[\frac{{{p}^{2}}{{t}^{2}}}{3m}\]

B)  \[\mu \,mg\sin \theta \]

C)   \[\mu \,mgcos\theta +mg\sin \theta \]

D)   \[\mu \,mgcos\theta -mg\sin \theta \]

Correct Answer: D

Solution :

                The free body diagram showing the various force acting on the block are as follows The fractional force \[{{(Kg)}^{-2}}\] acts opposite to direction of motion of block, given by \[\Delta {{\mathbf{p}}_{1}}=-\Delta {{\mathbf{p}}_{2}}\] where R is reaction of the plane on block                 \[\Delta E=-\Delta (PE+KE)=0\]                 \[\mathbf{F}\Delta t=m\Delta \mathbf{v}\]                 \[\Delta \mathbf{x}\propto \Delta \mathbf{F}\] Resultant force upwards                 \[1:8\]