JEE Main & Advanced Physics NLM, Friction, Circular Motion Work Done Against Friction

Work Done Against Friction (1) Work done over a rough inclined surface If a body of mass m is moved up on a rough inclined plane through distance s, then \[\operatorname{Work} done = force\,\,\times \,\,distance\]                            \[= ma\,\,\times \,\,s\]                         \[= mg [sin\,\theta ~+m\,cos\,\theta ]\]                         \[=mg\,s\,[\sin \theta +\mu \cos \theta ]\] (2) Work done over a horizontal surface In the above expression if we put \[\theta = 0\] then \[\operatorname{Work} done = force\,\,\times \,\,distance\]                            \[= F\,\,\times \,\,s\]                            \[=\,\,\mu mg s\] It is clear that work done depends upon (i) Weight of the body. (ii) Material and nature of surface in contact. (iii) Distance moved. Sample problems based on work done against friction Problem 15. A body of mass 5kg rests on a rough horizontal surface of coefficient of friction 0.2.  The body is pulled through a distance of 10m by a horizontal force of 25 N.  The kinetic energy acquired by it is \[\left( g = 10 m{{s}^{2}} \right)\] [EAMCET (Med.) 2000]             (a) 330 J            (b) 150 J            (c) 100 J            (d) 50 J             Solution: (b) Kinetic energy acquired by body = Total work done on the body ? Work done against friction             \[\operatorname{F}\,\,\times \,\,S-\mu mg\,S\,= 25 \times  10  0.2 \times  5 \times  10 \times 10 = 250  100 = 150 J.= 25 \times  10  0.2 \times  5 \times  10 \times 10 = 250  100 = 150 J.\] Problem 16. 300 Joule of work is done in sliding a 2 kg. block up an inclined plane to a height of 10 meters.  Taking value of acceleration due to gravity ?g? to be \[10 m/{{s}^{2}}\], work done against friction is [MP PMT 2002]             (a) 100 J            (b) 200 J            (c) 300 J            (d) Zero             Solution: (a) Work done against gravity =\[~mgh = 2 \times  10 \times  10 = 200J\] Work done against friction = Total work done ? Work done against gravity \[= 300  200 = 100J.\] Problem 17. A block of mass 1 kg slides down on a rough inclined plane of inclination \[{{60}^{o}}\] starting from its top. If the coefficient of kinetic friction is 0.5 and length of the plane is 1 m, then work done against friction is \[\left( Take g = 9.8 m/{{s}^{2}} \right)\] [AFMC 2000; KCET (Engg./Med.) 2001] (a) 9.82 J           (b) 4.94 J           (c) 2.45J            (d) 1.96 J Solution: (c) \[W=\mu mg\cos \theta .S=0.5\times 1\times 9.8\times \frac{1}{2}=2.45\text{ }J.\] Problem 18. A block of mass 50 kg slides over a horizontal distance of 1 m.  If the coefficient of friction between their surfaces is 0.2, then work done against friction is [CBSE PMT 1999, 2000; AIIMS 2000; BHU 2001] (a) 98 J             (b) 72J              (c) 56 J              (d) 34 J Solution: (a) \[W=\mu mgS=0.2\times 50\times 9.8\times 1=98J\].