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[OLYMPIADTEXTBOOK] IX - PHYSICS (Vol-1) 1 WORK, POWER AND ENERGY SYNOPSIS - 1 Work: In practical sense, we attach the phrase “work” with tiredness. Whenever we feel tired, (physically or mentally) we say, that we have done a lot of work, However, the concept of work in physics, is slightly different. When an applied force, happens to displace the point of application in the direction of the force work is said to have been done. Work done by a constant force: Let a constant force F  , displace a particle from point A to point B, as shown in the figure; then work done by the force is defined as the product of the force (magnitude) and the displacement (magnitude) of its point of application along the direction of the applied force. F S A M B  Thus work done W = F scos  [ the displacement AM of the particle along the applied force   ABcos ] [ where s  is the actual displacement and  , the angle between force and displacement]  W F.s    [Vectorially] Is S.I. system of measurement, force is expressed in newton, displacement in meters, and work done is obtained in joule. Note : Even though work done be expressed as N.m (dimensionally which is same as joule), but it is customary to use joule as the unit of work (or energy) and not newton–meter. Mathematically, we have W Fscos   If    0 ; s = 1m ; F = 1 N, then W = 1 J Thus, one joule can be defined as the work done by a constant force of 1 N, displacing the point of application by 1 m, along the direction of force. Regarding work done by force, it is worth, remembering that: (1) Since W Fscos    work done by a force can be positive or negative according as the value of cos  is positive or negative. (  F and s, being magnitudes, are always positive) F  F  S  S  A A   B B W = +ve for = acute angle  W = -ve for = obtuse angle 
[OLYMPIADTEXTBOOK] IX - PHYSICS (Vol-1) 2 Thus, if the displacement (of the point of application) has a component along the direction of applied force, then work done is positive. On the other hand, if the displacement has a component opposite to the direction of applied force, then work done will be negative. (2) If the applied force and particle’s displacement be mutually perpendicular, then work done by the force on the particle is zero. ( cos 90° being zero). Examples : (a) When a stone tied to one end of a string is whirled by holding the other end, then the tension (force) developed in the string being always normal to the direction of displacement of the stone, work done by the tension on the stone is zero. (See figure). s F O (b) When a vehicle successfully negotiates a rough horizontal curve, the work done by frictional forces on the vehicle is zero. O S  f  (c) A coolie with a luggage on his head, moving on a horizontal platform, does no work, since the direction of force is vertically up and displacement horizontal (even though he might feel physically tired). (Figure) N S (3) Since W Fscos   if s = 0 ; then W = 0 Note : if the point of application of force is unable to move, then work done by the force is zero. Example : If a boy tries to push a heavy boulder, by applying a force, but unable to displace it, then work done by the boy is zero. (see figure ).
[OLYMPIADTEXTBOOK] IX - PHYSICS (Vol-1) 3 (4) From W Fscos   If F = 0 ; W = 0. Thus, for work to be done, force is a must. (5) Since work done is the scalar (or dot) product of the force and displacement vectors, therefore, work is a scalar quantity. Total work done in several stages, can be obtained as the algebraic (scalar) sum of the work done at each stage. (6) If F  be a constant force acting on a particle, and the particle be displaced in several stages say, first from A to B, then from B to C, and then from C to D, with respective displacements s ,s 1 2   and s3  (Figure) s  F  s1  s2  s3  D C B A O Then total work done by the force F  is given by W F.s F.s F.s F. s s s 1 2 3 1 2 3                        [  Scalar product obeys distributive law]  W F.s    [where s  is supposed to be the total displacement ] (7) If a particle is being acted by constant forces F ,F ,F 1 2 3    etc. displacing it by s  , then work done by each force will be F .s, F .s, F .s 1 2 3       etc. Total work done by all forces W F .s F .s F .s ..... 1 2 3            W F F F ..... .s 1 2 3               = F.s   [where F  is the resultant of all forces] Thus, the total work done by individual forces is equal to the work done by their resultant. Work done due to a variable force : If the ‘direction’ or ‘magnitude’ or both, of a force, varies with the displacement of its point of application the small work done by the force during an elemental displacement dr  is found as F . dr   . [where F  is the force at the position r  of its point of application]. Next, these small works done, during the elemental displacements is summed up to get the total work done.
[OLYMPIADTEXTBOOK] IX - PHYSICS (Vol-1) 4 dr F r  X O Y P Thus total work done W = 2 1 r r F . dr    where the integration is to be carried out between the desired limits. Note : If F  is a constant force (that does not vary with r) then F  can be taken out of the integral sign. Thus W = F  . 2 1 r r dr     F . r r  2 1     which we have already discussed. To find the work done from force versus displacement curve : Given below shows the variation of a typical force ‘F’ with respect to a change in the displacement x of its point of application. X F X1 X2 dx X F Let us consider a thin strip of thickness dx, at a distance x from the force axis .Then the area of the strip = F dx [where F is the force at a displacement x] Now, the sum total of area between the ordinates at x = x1 and x = x2 and within the F – x curve and x - axis can be obtained as Area = 2 1 x x F dx  . But 2 1 x x F dx  is the work done by the force during the displacement of the point of application from x1 to x2 .  Work done between displacement x1 to x2 = Area below the F – x curve between the ordinates at x1 and x2