question_answer 75)

                  When a body slides down from rest along a smooth inclined plane making an angle of 45 0 with the horizontal, it takes time T. When die same body slides down from rest along a rough inclined plane making the same angle and through die same distance, it is seen to take time pT, where p is some number greater than 1. Calculate the coefficient of friction between the body and the rough plane.

Answer:

                  \[{{S}_{1}}=\frac{1}{2}\,{{a}_{1}}t_{1}^{2}\] and \[{{S}_{2}}=\frac{1}{2}\,{{a}_{2}}t_{2}^{2}\]                 Since \[{{S}_{1}}={{S}_{2}}\]                 \[\therefore \]\[{{a}_{2}}t_{2}^{2}\] or\[\frac{{{a}_{1}}}{{{a}_{2}}}\,=\,{{\left( \frac{{{t}_{2}}}{{{t}_{1}}} \right)}^{2}}\]                 Since \[{{t}_{1}}=T\] and \[{{t}_{2}}\,=\,pT\]                 \[\therefore \] \[\frac{{{a}_{1}}}{{{a}_{2}}}={{p}^{2}}\]                                 \[{{a}_{1}}=\,g\,\sin \,{{45}^{o}}\,=\frac{g}{\sqrt{2}}\]                 and \[{{a}_{2}}=g\,\sin \,{{45}^{o}}-\mu g\,\cos {{45}^{o}}\]                 \[=\frac{g}{\sqrt{2}}\,-\,\frac{\mu g}{\sqrt{2}}=\,\frac{g(1-\,\mu )}{\sqrt{2}}\]                 From eqn. (i) \[\frac{g}{g(1-\mu )}\,={{p}^{2}}\]                 or \[\mu =\,1-\,\frac{1}{{{p}^{2}}}\]