JEE Main & Advanced Physics NLM, Friction, Circular Motion Stopping of Block Due to Friction

Stopping of Block Due to Friction (1) On horizontal road (i) Distance travelled before coming to rest: A block of mass m is moving initially with velocity u on a rough surface and due to friction it comes to rest after covering a distance S. Retarding force \[F=ma=\mu R\]                         \[\Rightarrow \,\,\,ma=\mu \,mg\] \[\therefore \,\,\,\,\,\,\,\,a=\mu g\] From \[{{\operatorname{v}}^{2}}={{u}^{2}}-2aS\Rightarrow \,\] \[0={{u}^{2}}-2\mu \,g\,S\,\]             \[\text{ }\!\![\!\!\text{ As}\,\,v=0,\,\,a=\mu g]\] \[\therefore \,\,\,\,\,\,S=\frac{{{u}^{2}}}{2\mu g}\] or      \[S=\frac{{{P}^{2}}}{2\mu {{m}^{2}}g}\]    [As momentum P = mu] (ii) Time taken to come to rest From equation \[v=u-a\,t\Rightarrow \] \[0=u-\mu \,g\,t\]     \[[\text{As}\,\,\,v=0,\,a=\mu \,g]\] \[\therefore \,\,\,t=\frac{u}{\mu g}\] (iii) Force of friction acting on the body We know,          F = ma So,                   \[F=m\frac{(v-u)}{t}\]                                                                            \[F=\frac{mu}{t}\]                     [As v = 0]                         \[F=\mu \,mg\]               \[\left[ \text{As}\,\,t=\frac{u}{\mu g} \right]\] (2) On inclined road : When block starts with velocity u its kinetic energy will be converted into potential energy and some part of it goes against friction and after travelling distance S it comes to rest i.e. v = 0. And we know that retardation \[a=g\,[\sin \theta +\mu \cos \theta ]\] By substituting the value of v and a in the following equation                         \[{{v}^{2}}={{u}^{2}}-2a\,S\]             \[\Rightarrow \,\,\,\,0={{u}^{2}}-2g\,[\sin \theta +\mu \cos \theta ]\,S\]             \[\therefore \,\,\,\,\,S=\frac{{{u}^{2}}}{2g\,(\sin \theta +\mu \cos \theta )}\] Sample problems based on motion of body on rough surface Problem 27. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10s.  Then the coefficient of friction is [AIEEE 2003]             (a) 0.01             (b) 0.02             (c) 0.03             (d) 0.06 Solution: (d) \[v=u-at\] \[=u-\mu g\,t=0\]             \[\therefore \,\,\,\,\,\,\mu =\frac{u}{gt}=\frac{6}{10\times 10}=0.06\] Problem 28. A 2 kg mass starts from rest on an inclined smooth surface with inclination \[{{30}^{o}}\] and length 2 m.  How much will it travel before coming to rest on a surface with coefficient of friction 0.25 [UPSEAT 2003]             (a) 4 m             (b) 6 m              (c) 8 m              (d) 2 m Solution: (a) \[{{v}^{2}}={{u}^{2}}+2aS\]\[=0+2\times g\sin 30\times 2\]             \[v=\sqrt{20}\]             Let it travel distance ?S? before coming to rest              \[S=\frac{{{v}^{2}}}{2\mu g}=\frac{20}{2\times 0.25\times 10}=4m.\]