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  • A block of mass slides down on a rough inclined plane with constant velocity. The angle made by the i

A block of mass \mathrm{M} slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is \mathrm{\theta}. The magnitude of the contact force will be:
 

Option: 1

\mathrm{Mg}


Option: 2

\mathrm{Mg} \cos \theta


Option: 3

\sqrt{\mathrm{Mg} \sin \theta+\mathrm{Mg} \cos \theta}


Option: 4

\mathrm{Mg} \sin \theta \sqrt{1+\mu}


Answers (1)

best_answer

For block moving at constant velocity

\mathrm{\sum F_{x}=0}

\mathrm{-mg\sin \Theta +\mu Mg\cos \Theta =0}

\mathrm{\tan \Theta =\mu }

\mathrm{f=Mg\sin \Theta =\mu Mg\cos \Theta \rightarrow (1)}

\mathrm{Contact \: force=\sqrt{f^{2}+N^{2}}}

                            \mathrm{\sqrt{\left ( Mg\sin \Theta \right )^{2}+\left ( Mg\cos \Theta \right )^{2}}}

\mathrm{Contact \: force=Mg}

Hence 1 is correct option.

Posted by

Divya Prakash Singh

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