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Tell me? - Laws of motion - JEE Main

A smooth block is released at rest on a  45o incline and then slides a distance d . The time taken to slide  is  n time as much to slide on rough incline than on a smooth incline. The coefficient of friction is

  • Option 1)

    \mu _{s}= 1-\frac{1}{n^{2}}

  • Option 2)

    \mu _{s}= \sqrt{1-\frac{1}{n^{2}}}

  • Option 3)

    \mu _{k}= 1-\frac{1}{n^{2}}

  • Option 4)

    \mu _{k}= \sqrt{1-\frac{1}{n^{2}}}

 
Answers (1)
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As we learnt in

Kinetic or Dynamic Friction -

f_{K}\;\alpha\ R

f_{K}=\mu_{K} R

f_{K}= kinetic friction 

\mu_{K}= coefficient of kinetic friction

R = reaction

- wherein

f_{K}<F_{l}

\therefore\ \mu_{K}<\mu_{s}

\mu_{K}=depends on the nature of surface in contact.

 

 

 

Component of g down the plane =\; gsin\theta

\therefore      For smooth plane,

d=\frac{1}{2}(gsin\theta )t^{2}\; ............(i)

For rough plane,

Frictional retardation up the plane =\; \mu _{k}(gcos\theta )

\therefore \; \; \; d=\frac{1}{2}(gsin\theta -\mu _{k}gcos\theta )(nt)^{2}

\therefore \; \; \; \frac{1}{2}(gsin\theta )t^{2}=\frac{1}{2}(gsin\theta -\mu _{k}gcos\theta )n^{2}t^{2}

or     sin\theta =n^{2}(sin\theta -\mu _{k}cos\theta )

Putting \theta =45^{\circ}

or \; \; sin45^{\circ}=n^{2}(sin45^{\circ}-\mu _{k}cos45^{\circ})

or\; \; \; \frac{1}{\sqrt{2}}=\frac{n^{2}}{\sqrt{2}}(1-\mu _{k})

or\; \; \; \mu _{k}=1-\frac{1}{n^{2}}

Correct option is 3.

 

 


Option 1)

\mu _{s}= 1-\frac{1}{n^{2}}

This is an incorrect option.

Option 2)

\mu _{s}= \sqrt{1-\frac{1}{n^{2}}}

This is an incorrect option.

Option 3)

\mu _{k}= 1-\frac{1}{n^{2}}

This is the correct option.

Option 4)

\mu _{k}= \sqrt{1-\frac{1}{n^{2}}}

This is an incorrect option.

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