Stopping of Block Due to Friction
Category : JEE Main & Advanced
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 \[S=\frac{{{v}^{2}}}{2\mu g}=\frac{20}{2\times 0.25\times 10}=4m.\]
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