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A body starts sliding down from the top of an inclined plane at an angle theta with the horizontal direction. The first one third of the incline is smooth, the next one third has coefficient of friction mu/2 and the last one third has coefficient of friction mu. If the body comes to rest at the bottom of the plane then the value of mu is |
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Answer» `(tantheta)/2`  Let mass of body = m and AB= BC = CD = x [Given in AB, BC and CD plane coefficient of friction respectively 0, `mu/2 " and "mu`] For inclined plane AB, friction force, F = 0 So, the net force equation, `mgsintheta-0=ma` or `a=gsintheta` Here, from THIRD equation of the motion, `""v^(2)=u^(2)+2ax` or `""v^(2)=2 times gsintheta times x=2gxsintheta[because u=0]` For inclined plane BC, friction force, `F=(mu//2)mgcostheta` So, net force equation, `rArr mgsintheta-(mu//2)mgcostheta=ma^(.)` `rArr a^(.)=gsintheta-mu/2gcostheta` From third equation of the motion, `v^(2)=u^(2)+2a^(.)x` `rArr v^(2)=u^(2)+2a^(.)x ""[because " inthis case " u^(.)=v]` `rArr v^(2)=2gxsintheta+2(gsintheta-mu/2gcostheta)x` `""v^(2)=4gxsintheta-mugxcostheta` For inclined plane CD, friction force, `F=mumgcostheta` So, net force equation, `rArr mgsintheta-mumgcostheta=ma^(.)` `rArr a^(.)=gsintheta-mugcostheta` From third equation of the motion, `rArr v..=mu^(.)+2a^(.)x=mu^(.)+2a^(.)x ""(because mu^(.)=v^(.))` `rArr 0=4gxsintheta-mugxcostheta+2(gsintheta-mugcostheta)x` `rArr 0=6 gxsintheta=3mugx costheta rArr` 6gxsintheta=3ugxcostheta` `rArr mu=(6gxsintheta)/(3gxcostheta) rArr mu=2tantheta` |
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