JEE Main & Advanced Physics NLM, Friction, Circular Motion Acceleration of Block

Acceleration of Block (1) Acceleration of a block on horizontal surface When body is moving under application of force P, then kinetic friction opposes its motion. Let a is the net acceleration of the body From the figure             \[ma=P-{{F}_{k}}\] \[\therefore \,\,\,\,\,\,\,a=\frac{P-{{F}_{k}}}{m}\] (2) Acceleration of a block down a rough inclined plane             When angle of inclined plane is more than angle of repose, the body placed on the inclined plane slides down with an acceleration a. From the figure         \[ma=mg\sin \theta -F\] Þ                           \[ma=mg\sin \theta -\mu R\] Þ                          \[ma=mg\sin \theta -\mu \,mg\,\,\cos \theta \]    \[\therefore \] Acceleration \[\therefore \,\,\,\,\,\,\,\,a=g\,[\sin \theta -\mu \cos \theta ]\] Note: q For frictionless inclined plane \[\mu =0\] \[\therefore a=g\sin \theta \]. (3) Retardation of a block up a rough inclined plane             When angle of inclined plane is less than angle of repose, then for the upward motion                         \[ma=mg\,\,\sin \theta +F\]             \[ma=mg\,\,\sin \theta +\mu \,mg\,\,\cos \theta \]             Retardation \[a=g\,[\sin \theta +\mu \,\cos \theta ]\]             Note: q For frictionless inclined plane \[\mu =0\] \[\therefore a=g\,\sin \theta \] Sample problems based on acceleration against friction Problem 11. A body of mass 10 kg is lying on a rough plane inclined at an angle of 30o to the horizontal and the coefficient of friction is 0.5. The minimum force required to pull the body up the plane is [JIPMER 2000]             (a) 914 N           (b) 91.4 N          (c) 9.14 N          (d) 0.914 N             Solution: (b) \[F=mg(\sin \theta +\mu \cos \theta )\]\[=10\times 9.8\,(\sin 30+0.5\,\cos 30)=91.4\,N\] Problem 12. A block of mass 10 kg is placed on a rough horizontal surface having coefficient of friction \[\mu = 0.5\]. If a horizontal force of 100 N is acting on it, then acceleration of the block will be [AIIMS 2002]             (a)  \[0.5 m/{{s}^{2}}\] (b) \[5 m/{{s}^{2}}\]        (c) \[10 m/{{s}^{2}}\]   (d) \[15 m/{{s}^{2}}\]             Solution: (b) \[a=\frac{\text{Applied }\,\text{force -- kinetic}\,\text{friction}}{\text{mass}}\]\[=\frac{100-0.5\times 10\times 10}{10}\]\[= 5m/{{s}^{2}}.\] Problem 13. A body of weight 64 N is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards.  If the coefficients of static and dynamic friction are 0.6 and 0.4 respectively, the acceleration of the body will be (Acceleration due to gravity = g) [EAMCET 2001]             (a) \[\frac{g}{6.4}\]        (b) 0.64 g           (c) \[\frac{g}{32}\]                     (d) 0.2 g             Solution: (d) Limiting friction \[={{F}_{1}}={{\mu }_{s}}R\Rightarrow \,\,64=0.6\text{ }m~g\Rightarrow m=\frac{64}{0.6\,g}\]             \[Acceleration=\frac{\text{Applied force -- Kinetic}\,\,\text{ friction}}{\text{Mass of the body}}=\] \[\frac{64-{{\mu }_{K}}mg}{m}\]\[=\frac{64-0.4\times \frac{64}{0.6}}{\frac{64}{0.6g}}\,\,=\,\,0.2\,g\] Problem 14. If a block moving up at \[\theta ={{30}^{o}}\] with a velocity 5 m/s, stops after 0.5 sec, then what is \[\mu \] [CPMT 1995]                         (a) 0.5               (b) 1.25                          (c) 0.6               (d) None of these             Solution: (c) From \[v=u-at\Rightarrow 0=u-at\] \[\therefore \,\,\,\,t=\frac{u}{a}\]             for upward motion on an inclined plane \[a=g(\sin \theta +\mu \cos \theta )\] \[\therefore \,\,\,\,\,\,\,\,t=\frac{u}{g(\sin \theta +\mu \cos \theta )}\]             Substituting the value of \[\theta ={{30}^{o}},\,t=0.5\,\sec \] and \[u=5m/s\], we get \[\mu =0.6\]