Question

Object takes "n" times more to slide down a 45 degree rough incline plane in comparison to slides down a perfectly smooth 45 degre inlcine.The coeff of kinetic friction between object and incline is ????????????????????????????

Answer 1

Hey there !!!!!

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On smooth inclined plane acceleration of the body:

                        a=gsinθ

If "s" is distance traveled , u=0 and t₁ is time taken 

                    s=ut+at₁²/2

                    s=0+gt₁²sinθ/2

                   s=gt₁²sinθ/2-------Equation 1


On rough inclined plane if "p" is coefficient of friction acceleration is

                                   a=gsinθ- pgcosθ

Distance traveled if u=0   time=t₂

                     s=ut₂+at₂²/2

                    s=0+(gsinθ- pgcosθ)t₂²/2

                    s=(gsinθ- pgcosθ)t₂²/2----Equation 2

As the body travels the same distance irrespective of nature of the surface

Equating equation 1 & 2

                             gt₁²sinθ/2=(gsinθ- pgcosθ)t₂²/2

                             gt₁²sinθ=(gsinθ- pgcosθ)t₂²

                         t₂²/t₁²=gsinθ/(gsinθ- pgcosθ)

But according to question t₂=nt₁

                         t₂²/t₁²=n²

                               n²=gsinθ/(gsinθ- pgcosθ)

                              n²=sinθ/sinθ-pcosθ

                             n²(sinθ-pcosθ)=sinθ

                             n²sinθ-n²pcosθ=sinθ

                             n²sinθ-sinθ=n²pcosθ

                            sinθ(n²-1)/cosθ*n²=p

                           sinθ/cosθ=tanθ

So,
                            tanθ(n²-1)/n²=p

 But  θ=45 and tan45=1

So          tan45(n²-1)/n²=p

                           p=n²/n²-1/n²

                              p=1-1/n²

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Hope this helped you...................



                             


                   

Answer 2

Answer:

coefficient of kinetic friction is (1 - 1/n^2).

this is the correct answer