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The upper half of an inclined plane of inclination \theta is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by:

  • Option 1)

    \mu =\frac{2}{tan\theta}

  • Option 2)

    \mu=2 tan\theta

  • Option 3)

    \mu=tan\theta

  • Option 4)

    \mu=\frac{1}{tan\theta}

 

Answers (1)

As we learnt in 

Work done over a rough inclined surface -

mg=[sin heta+mu cos heta]

mg = weight

 

- wherein

W=F imes displacement

S= displacement

 

 

 

Suppose length of plane is L

\Delta K=\Delta U

mgL\sin \Theta ----\left ( 1 \right )

work done by friction

\mu \left ( mg \cos \Theta \right )\times \frac{1}{2}       (work done by friction for upper half is 0)

W.D=\Delta K

mgL\sin \Theta =\left ( mg\cos \Theta \right )\mu \times \frac{L}{2}

\tan \Theta =\frac{\mu}{2}

\mu=2\tan \Theta

Correct option is 2.


Option 1)

\mu =\frac{2}{tan\theta}

Incorrect

Option 2)

\mu=2 tan\theta

Correct

Option 3)

\mu=tan\theta

Incorrect

Option 4)

\mu=\frac{1}{tan\theta}

Incorrect

Posted by

Vakul

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