Work, Energy and Power

Work, Energy and Power INTRODUCTION The term ‘work’ as understood in everyday life has a different meaning in scientific sense. If a coolie is carrying a load on his head and waiting for the arrival of the train, he is not performing any work in the scientific sense. In the present study, we shall have a look into the scientific aspect of this most commonly used term i.e., work. WORK DONE BY CONSTANT FORCE The physical meaning of the term work is entirely different from the meaning attached to it in everyday life. In everyday life, the term ‘work’ is considered to be synonym of ‘labour’, ‘toil’, effort’ etc. In physics, there is a specific way of defining work. Work is said to be done by a force when the force produces a displacement in the body on which it acts in any direction except perpendicular to the direction of the force. For work to be done, following two conditions must be fulfilled. (i) A force must be ap plied. (ii) The applied force must produce a displacement in any direction except perpendicular to the direction of the force.  Suppose a force F is applied on a body in such a  way that the body suffers a displacement S in the direction of the force. Then the work done is given by W = FS However, the displacement does not always take place in the direction of the force. Suppose a constant  force F , applied on a body, produces a displacement  S   in the body in such a way that S is inclined to F at an angle . Now the work done will be given by the dot product of force and displacement.   W= F . S Since work is the dot product of two vectors therefore it is a scalar quantity. W = FS cos  or W = (F cos )S  W = component of force in the direction of displacement × magnitudes of displacement. So work is the product of the component of force in the direction of displacement and the magnitude of the displacement. Also, W = F (S cos ) or work is product of the component of displacement 3. in the direction of the force and the magnitude of the displacement. Work, Energy and Power The tension in the string of a simple pendulum is always perpendicular to displacement. (Figure). So, work done by the tension is zero. Case (ii) : When S = 0, then W = 0. So, work done by a force is zero if the body suffers no displacement on the application of a force. Example : Special Cases : Case (i) When  = 90º, then W = FS cos 90º = 0 A person carrying a load on his head and standing at So, work done by a force is zero if the body is a given place does no work. displaced in a direction perpendicular to the Case (iii) : direction of the force. When 0º   < 90º [Figure], then cos  is Examples : positive. Therefore. 1. Consider a body sliding over a horizontal surface. W (= FS cos ) is positive. The work done by the force of gravity and the reaction of the surface will be zero. This is because Work done by a force is said to be positive if both the force of gravity and the reaction act normally the applied force has a component in the to the displacement. direction of the displacement. 2. Examples : 1. When a horse pulls a cart, the applied force and the displacement are in the same direction. So, work done by the horse is positive. The same argument can be applied to a man carrying 2. When a load is lifted, the lifting force and the displacement act in the same direction. So, work a load on his head and walking on a railway platform. done by the lifting force is positive. Consider a body moving in a circle with constant speed. At every point of the circular path, the 3. When a spring is stretched, both the stretching force and the displacement act in the same direction. So, centripetal force and the displacement are mutually work done by the stretching force is positive. perpendicular (Figure). So, the work done by the centripetal force is zero. The same argument can be applied to a satellite moving in a circular orbit. In Case (iv) : When 90º <   180º (Figure), then cos is this case, the gravitational force is always negative. Therefore W (= FS cos ) is negative. perpendicular to displacement. So, work done by gravitational force is zero. Work done by a force is said to be negative if the applied force has component in a direction opposite to that of the displacement. Examples : 1. When brakes are applied to a moving vehicle, the work done by the braking force is negative. This is because the braking force and the displacement act in opposite directions. Work, Energy and Power When a body is dragged along a rough surface, the So, the work done on the particle is the sum of the work done by the frictional force is negative. This is individual works done by all the forces acting on the because the frictional force acts in a direction particle. opposite to that of the displacement. Important points about work: 3. When a body is lifted, the work done by the 1. Work is defined for an interval or displacement. gravitational force is negative. This is because the There is no term like instantaneous work similar to gravitational force acts vertically downwards while instantaneous velocity. the displacement is in the vertically upwards 2. For a particular displacement, work done by a force direction. is independent of type of motion i.e. whether it moves with constant velocity, constant acceleration or retardation etc. Ex.1 Figure shows four situations in which a force acts 3. For a particular displacement work is independent on a box while the box slides rightward a distance d of time. Work will be same for same displacement across a frictionless floor. The magnitudes of the whether the time taken is small or large. forces are identical, their orientations are as shown. 4. When several forces act, work done by a force for Rank the situations according to the work done on a particular displacement is independent of other the box during the displacement, from most positive forces. to most negative. 5. A force is independent from reference frame. Its displacement depends on frame so work done by a force is frame dependent therefore work done by a (A) (B) force can be different in different reference frame. 6. Effect of work is change in kinetic energy of the particle or system. (C) (D) 7. Work is done by the source or agent that applies the Ans. D, C, B, A force. 2. Explanation : UNITS OF WORK In (D),  = 0º, cos  = 1 (maximum value). So, 1. Unit of work : work done is maximum. I. In cgs system, the unit of work is erg. In (C),  = 90º, cos  is positive. Therefore, W is One erg of work is said to be done when a force of positive. one dyne displaces a body through one centimetre In (B),  = 90º, cos  is zero. W is zero. in its own direction. In (A),  is obtuse, cos  is negative. W is negative.  1 erg = 1 dyne × 1 cm = 1g cm s–2 × 1 cm = 1 g cm2 s–2 WORK DONE BY MULTIPLE FORCES Note, Erg is also called dyne centimetre. If several forces act on a particle, then we can replace II. In SI i.e., International System of units, the unit of     in equation W = . by the net force where  F S F F work is joule (abbreviated as J). It is named after     the famous British physicist James Personal Joule  F = F 1 + F 2 + F3 + .....   (1818 – 1869).  W = [ F ] . S ...(i) One joule of work is said to be done when a force This gives the work done by the net force during a of one newton displaces a body through one metre  displacement S of the particle. in its own direction. We can rewrite equation (i) as : 1 joule = 1 newton × 1 metre = 1 kg ×1 m/s2 × 1 m       = 1kg m2 s–2 W = F 1 . S + F 2 . S + F 3 . S + ..... Note Another name for joule is newton metre. or W = W + W + W + .......... 1 2 3 Relation between joule and erg 1 joule = 1 newton × 1 metre 1 joule = 105 dyne × 102 cm = 107 dyne cm 1 joule = 107 erg 1 erg = 10–7 joule Work, Energy and Power Sol. Force, F = 50 N ; Distance, S = 100 m ; Angle ,  = 60º DIMENSIONS OF WORK W = FS cos  = 50 × 100 × cos 60º joule [Work] = [Force] [Distance] 1 = [MLT–2] [L] W = 50 × 100 × J = 2500 J 2 = [ML2T–2] 1 Work has one dimension in mass, two dimensions in [ cos 60º = ] 2 length and ‘–2’ dimensions in time, 2500 On the basis of dimensional formula, the unit of work Wages = × 10 paise = 10 rupees 2 –2 25 is kg m s . Note that 1 kg m2 s–2 = (1 kg m s–2) m = 1 N m = 1 J. Ex.4 Calculate the work done in raising a stone of mass 5 kg and specific gravity 3 lying at the bed of a lake through a height of 5 metre. Ex.2 A cyclist comes to a skidding stop in 10 m. During Sol. When a body is immersed in water, its apparent weight is decreased in accordance with the this process, the force on the cycle due to the road Archimedes’ principle. is 200 N and is directly opposite to the motion. weight in air 5 kg wt (a) How much work does the road do on the cycle? Loss of weight in water = specific gravity = 3 (b) How much work does the cycle do on the road? 5  10  Weight of stone in water =  5  3  kg wt = kg wt 3   Sol. (a) The work done on the cycle by the road is the 10 10 98 work done by the frictional force exerted by the road Force, F = kg wt = × 9.8 N = N 3 3 3 on the cycle.   98 W = F . S = FS cos 180º Work done, W = × 5 J = 163.3 J. 3 or W = – FS or W = – 200 N × 10 m WORK IN TERMS OF or W = – 2000 J RECTANGULAR COMPONENTS It is this negative work which brings the cycle to   In terms of rectangular components, and may S F rest. This is clearly in accordance with work-energy be written as : theorem.   (b) The displacement of the road is zero. So, work F = Fx î + Fy ĵ + Fz k̂ and S = Sx î + Sy ĵ + Sz k̂ done by the cycle on the road is zero.   F . S = (Fx î + Fy ĵ + Fz k̂ ) .(Sx î + Sy ĵ + Sz k̂ ) (Using Newton’s third law of motion, an equal and opposite force acts on the road due to the cycle. = F x S x ( î . î ) + F x S y ( î . ĵ ) + F x S z ( î . k̂ ) + The magnitude of this force is 200 N.) F y S x ( ĵ . î ) + F y S y ( ĵ . ĵ ) + F y S z ( ĵ . k̂ ) + Ex.3 A gardener moves a lawn roller through a distance FzSx( k̂ . î ) + FzSy(k̂ . ĵ ) + FzSz( k̂ . k̂ ) of 100 metre with a force of 50 newton. Calculate his wages if he is to be paid 10 paise for doing 25 But î . ĵ = î . k̂ = ĵ . î = ĵ . k̂ = k̂ . î = k̂ . ĵ = 0 joule of work. It is given that the applied force is   î . î = ĵ . ĵ = k̂ . k̂ = 1 [ F. S = FxSx + Fy Sy + FzSz] inclined at 60º to the direction of motion. Now, Work, Energy and Power The total work done will be sum of infinitely small work Ex.5 A body constrained to move along the z-axis of a coordinate system is subjected to a constant force B   B   W A B  F.ds  (F cos  )ds   F given by  A  F = – î + 2 ĵ +3 k̂ N A In terms of rectangular components,  F where î , ĵ , k̂ are unit vectors along the x, y  = Fx î + Fy ĵ + Fz k̂ ds = dx î + dy ĵ + dz k̂ xB and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4m along the z-axis? W A B  yB F x x  F dy   F dz dx  A zB y y z z A A Sol. Since the body is displaced 4 m along z-axis only,   S Also,  F = – î + 2 ĵ +3 k̂ = 0 î + 0 ĵ + 4 k̂ Ex. 7 An object is displaced from position vector  r1  ( 2 î  3 ĵ) m  F  (3 x 2 î  2 y ĵ) N . Find the work done by this force.   Work done, W= F . S = (– î + 2 ĵ +3 k̂ ). (0 î + 0 ĵ + 4 k̂ ) = 12( k̂ . k̂ ) joule = 12 joule. Ex. 6 An object is displaced from point A(2m, 3m, 4m) to a point B(1m, 2m, 3m) under a constant force  F  (2 î  3 ĵ  4k̂ ) N . Find the work done by this force in this process. Sol. W =    rf  ri  to r2  ( 4 ĵ  6k̂ ) m under a force Sol. W =   rf  ri    F  dr  r2 2 =  r (3 x î  2 yĵ)  (dx î  dy ĵ  dzk̂ ) 1  r2  r1 (3 x 2 dx  2ydy ) = [ x 3  y 2 ] (( 24,, 36)) = 83 J AREA UNDER FORCE   F  dr DISPLACEMENT CURVE (1m, 2m, 3m ) ( 2 î  3 ĵ  4k̂ )  ( dx î  dy î  dzk̂ ) Graphically area under the force-displacement is the work done ( 2m, 3m, 4m ) = [2x  3 y  4z]((21mm,, 32mm,43mm)) =–9J Ans. WORK DONE BY A VARIABLE FORCE When the magnitude and direction of a force vary in three dimensions, it can be expressed as a function of the position. For a variable force work is calculated for infinitely small displacement and for this displacement force is assumed to be constant   dW = F.ds The work done can be positive or negative as per the area above the x-axis or below the x-axis respectively. Work, Energy and Power Ex.8 A force F = 0.5x + 10 acts on a particle. Here F is in newton and x is in metre. Calculate the work done by the force during the displacement of the particle from x = 0 to x = 2 metre. Sol. Small amount of work done dW in giving a small  displacement dx is given by dW = F . dx or dW = Fdx cos 0º or dW = Fdx [ cos 0º = 1] Total work done, W =  x 2 Fdx  x 0 =  x 2 x 0 0.5 x dx    x 2 x2 x 0 (0.5 x + 10)dx x 0 10 dx = 0.5 x2 2 x2 + 10 x 0 x 2 x W  (Force vector )  x 0 = 0 .5 2 “Basic concept of work lies in following lines” Draw the force at proper point where it acts that give proper importance to the point of application of force. Think independently for displacement of point of application of force, Instead of relating the displacement of applicant point with force relate it with the observer or reference frame in which work is calculated. [22 – 02] + 10[2 – 0] = (1 + 20) = 21 J  displacement vector of point of    application of     force as seen by observer   INTERNAL WORK Suppose that a man sets himself in motion backward by pushing against a wall. The forces acting on the ENERGY Definition: Energy is defined as internal capacity man are his weight ‘W’, the upward force N exerted of doing work. When we say that a body has energy by the ground and the horizontal force N´ exerted we mean that it can do work. by the wall. The works of ‘W’ and of N are zero because they are perpendicular to the motion. The Energy appears in many forms such as mechanical, force N´ is the unbalanced horizontal force that electrical, chemical, thermal (heat), optical (light), imparts to the system a horizontal acceleration. The acoustical (sound), molecular, atomic, nuclear etc., work of N´, however, is zero because there is no and can change from one form to the other. motion of its point of application. We are therefore confronted with a curious situation in which a force KINETIC ENERGY is responsible for acceleration, but its work, being Definition: zero, is not equal to the increase in kinetic energy of Kinetic energy is the internal capacity of doing work the system. of the object by virtue of its motion. The new feature in this situation is that the man is a Kinetic energy is a scalar property that is associated composite system with several parts that can move with state of motion of an object. An aeroplane in in relation to each other and thus can do work on straight and level flight has kinetic energy of each other, even in the absence of any interaction translation and a rotating wheel on a machine has with externally applied forces. Such work is called kinetic energy of rotation. If a particle of mass m is internal work. Although internal forces play no role moving with speed ‘v’ much less than the speed of in acceleration of the composite system, their points the light than the kinetic energy ‘K’ is given by of application can move so that work is done; thus 1 the man’s kinetic energy can change even though K  mv 2 2 the external forces do no work. Work, Energy and Power Important Points for K.E.   As mass m and v2 ( v . v ) are always positive, kinetic Ex.9 The kinetic energy of a body is increased by 21%. energy is always positive scalar i.e, kinetic energy What is the percentage increase in the magnitude of can never be negative. linear momentum of the body? The kinetic energy depends on the frame of reference, 121 121 1 1 Sol. Ek2 = Ek1 or m v22 = mv12 2 1. 2. 100 p K = 2 m and P = 2 m K ; P = linear momentum The speed v may be acquired by the body in any manner. The kinetic energy of a group of particles or bodies is the sum of the kinetic energies of the individual particles. Consider a system consisting of n particles of masses m1, m2, ......, mn. Let v 1 , v 2 , ..... , v n be their respective velocities. Then, the total kinetic energy Ek of the system is given by Ek = 1 2 m1v12 + 1 1 m v 2 + ......... + mnvn2 2 2 2 2 If n is measured in gram and v in cm s–1, then the kinetic energy is measured in erg. If m is measured in kilogram and v in m s–1, then the kinetic energy is measured in joule. It may be noted that the units of kinetic energy are the same as those of work. Infact, this is true of all forms of energy since they are interconvertible. 1 2 3 4 5 6 Object Mass (kg) –26 Air molecule 10 Rain drop at terminal speed 3.5 × 10–5 Stone dropped from 10 m 1 Bullet 5 × 10–5 Running athlete 70 Car 2000 Speed K(J) (m s–1) –21 500  10 9 1.4 × 10–3 14 10 2 200 10 3 10 3.5 × 103 25 6.3 × 105 RELATION BETWEEN MOMENTUM AND KINETIC ENERGY Consider a body of mass m moving with velocity v. Linear momentum of the body, p = mv Kinetic energy of the body, Ek = Ek = 1 2m (m2v2) or Ek = or v2 = 11 v 10 1 or mv2 = or p2 = 11 p 10 1 or p2 p1 –1= = 1 10 × 100 = 10 or p 2  p1 p1 × 100 11 10 mv1 11 10 –1= 1 10 So, the percentage increase in the magnitude of linear momentum is 10%. Ex.10 The linear momentum of a body is increased by 10%. What is the percentage change in its kinetic energy? Ans. Percentage increase in kinetic energy = 21%] 2  110 11 E1  11  121  H int .mv 2  mv 1, v 2  v 1,    100 10 E1  10  100  Percentage increase in kinetic energy   E1  E1  100  E1  = 21% Typical kinetic energies (K) S.No. 100 2 2  U2 dU   U1  r2 r1   F dr r2   i.e., U 2  U1    r F  d r   W 1 Whenever and wherever possible, we take the reference point at  and assume potential energy to be zero there, i.e., If we take r1 =  and U1 = 0 then 1 mv2 2 p2 or p = 2m POTENTIAL ENERGY Definition: Potential energy is the internal capacity of doing work of a system by virtue of its configuration. In case of conservative force (field) potential energy is equal to negative of work done by the conservative force in shifting the body from some reference position to given position. Therefore, in case of conservative force 2mE k U  r    F  d r  W Work, Energy and Power Important Points for P.E. : 1. Potential energy can be defined only for conservative forces. It has no relevance for non-conservative forces. 2. Potential energy can be positive or negative, depending upon choice of frame of reference. 3. Potential energy depends on frame of reference but change in potential energy is independent of reference frame. 4. Potential energy should be considered to be a property of the entire system, rather than assigning it to any specific particle. 5. It is a function of position and does not depend on the path. Types of Potential Energy (a) Elastic Potential Energy: It is the energy associated with state of compression or expansion of an elastic (spring like) object and is given by: MECHANICAL ENERGY Definition Mechanical energy ‘E’ of an object or a system is defined as the sum of kinetic energy ‘K’ and potential energy ‘U’, i.e., E=K+U Important Points for M.E.: 1. It is a scalar quantity having dimensions [ML2T-2] and SI units joule. 2. It depends on frame of reference. 3. A body can have mechanical energy without having either kinetic energy or potential energy. However, if both kinetic and potential energies are zero, mechanical energy will be zero. The converse may or may not be true, i.e., if E = 0 either both PE and KE are zero or PE may be negative and KE may be positive such that KE + PE = 0. 4. As mechanical energy E = K + U, i.e., E - U = K. Now as K is always positive, E - U  0,i.e., for 1 U  k y2 existence of a particle in the field, E  U. 2 where k is force constant and ‘y’ is the stretch or 5. As mechanical energy E = K + U and K is always compression. Elastic potential energy is always positive, so, if ‘U’ is positive ‘E’ will be positive. positive. However, if potential energy U is negative, ‘E’ will be (b) Electric Potential Energy: positive if K > |U| and E will be negative if K < |U| It is the energy associated with charged particles i.e., mechanical energy of a body or system can be that interact via electric force. For two point charges negative, and negative mechanical energy means that q1 and q2 separated by a distance ‘r’, potential energy is negative and in magnitude it is 1 q1q2 U more than kinetic energy. Such a state is called bound 4 0 r state, e.g., electron in an atom or a satellite moving As charge can be positive or negative, therefore around a planet are in bound state. electric potential energy can also be positive or negative. (c) Gravitational Potential Energy: Ex.11 A small block of mass 100 g is pressed against a It is due to gravitational force. For two particles of horizontal spring fixed at one end to compress the masses m1 and m2 separated by a distance ‘r’, it is spring through 5.0 cm (figure). The spring constant given by: is 100 N/m. When released, the block moves U  G m1m 2 r which for a body of mass ‘m’ at height ‘h’ relative to surface of the earth reduces to U = mgh Gravitational potential energy can be positive or negative. horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring ? Sol. When block released, the block moves horizontally with speed V till it leaves the spring. By energy conservation V2 = kx 2 m  V= Time of flight t = 1 2 kx 2 = 1 mv2 2 kx 2 m Work, Energy and Power CONSERVATIVE FORCES A force is said to be conservative if work done by or against the force in moving a body depends only on the initial and final positions of the body and not on the nature of path followed between the initial and final positions. 2H g So. horizontal distance travelled from the free end of the spring is V × t Consider a body of mass m being raised to a height h vertically upwards as show in above figure. The = work done is mgh. Suppose we take the body along the path as in (b). The work done during horizontal 22 100  (0.05 )2 motion is zero. Adding up the works done in the = × = 1m 10 0.1 two vertical parts of the paths, we get the result mgh once again. Any arbitrary path like the one shown in So, At a horizontal distance of 1 m from the free end (c) can be broken into elementary horizontal and of the spring. vertical portions. Work done along the horizontal Ex.12 A rigid body of mass m is held at a height H on parts is zero. The work done along the vertical parts two smooth wedges of mass M each of which are add up to mgh. Thus we conclude that the work themselves at rest on a horizontal frictionless floor. done in raising a body against gravity is independent On releasing the body it moves down pushing aside of the path taken. It only depends upon the initial the wedges. The velocity of recede of the wedges and final positions of the body. We conclude from from each other when rigid body is at a height h this discussion that the force of gravity is a from the ground is conservative force. Examples of Conservative forces. (i) Gravitational force, not only due to the Earth but in its general form as given by the universal law of gravitation, is a conservative force. (ii) Elastic force in a stretched or compressed spring is a conservative force. (iii) Electrostatic force between two electric charges is a conservative force. 2mg(H  h) 2mg(H  h) (iv) Magnetic force between two magnetic poles is a (A) (B) m  2M 2m  M conservative forces. Forces acting along the line joining the centres of 8mg ( H  h) 8mg ( H  h) (C) (D) two bodies are called central forces. Gravitational m  2M 2m  M force and Electrostatic forces are two important Ans. (C) examples of central forces. Central forces are conservative forces. kx 2 × m 2H g PROPERTIES OF CONSERVATIVE FORCES (i) Work done by or against a conservative force depends only on the initial and final positions of the body. (ii) Work done by or against a conservative force does no depend upon the nature of the path between initial and final positions of the body. If the work done a by a force in moving a body from an initial location to a final location is independent of the path taken between the two points, then the force is conservative. (iii) Work done by or against a conservative force in a round trip is zero. If a body moves under the action of a force that does no total work during any round trip, then the force is conservative; otherwise it is nonconservative. The concept of potential energy exists only in the case of conservative forces. (iv) The work done by a conservative force is completely recoverable. Complete recoverability is an important aspect of the work of a conservative force. Work, Energy and Power The quantity in parentheses is referred to as the scalar gradient of the function U and is denoted by grad U or U. We shall use the second, more convenient, designation where  (“nabla”) signifies the symbolic vector or operator     = i x  j y  k z Potential Energy curve (a) A graph plotted between the PE a particle and its displacement from the centre of force field is called PE curve. (b) Using graph, we can predict the rate of motion of a particle at various positions. (c) Force on the particle is F(x) = – dU dx Case : I On increasing x, if U increases, force is in CONSERVATIVE FORCE & (–) ve x direction i.e. attraction force. POTENTIAL ENERGY Case : II On increasing x, if U decreases, force is FS = –  U/  s, in (+) ve x-direction i.e. repulsion force. i.e. the projection of the field force, the vector F, at Different positions of a particle :a given point in the direction of the displacement dr Position of equilibrium : equals the derivative of the potential energy U with If net force acting on a body is zero, it is said to be in respect to a given direction, taken with the opposite sign. The designation of a partial derivative  /  s dU equilibrium. For equilibrium = 0. Points P, Q R emphasizes the fact of deriving with respect to a dx definite direction. and S are the states of equilibrium positions. So, having reversed the sign of the partial derivatives Types of equilibrium : of the function U with respect to x, y, z, we obtain (a) Stable equilibrium : When a particle is displaced the projection Fx, Fy and Fz of the vector F on the slightly from a position and a force acting on it brings unit vectors i, j and k. Hence, one can readily find it back to the initial position, it is said to be in stable the vector itself : F = Fxi + Fuj + Fzk, or equilibrium position. F=–  U U U   i j k y z  .  x Necessary conditions : – dU dx = 0, and d2U dx 2 = +ve Work, Energy and Power The velocity-dependent forces such as air resistance, (b) Unstable Equilibrium : When a particle is displaced slightly from a position and force acting on it tries to displace the particle further away from the equilibrium position, it is said to be in unstable equilibrium. Condition : – i.e. = d2U dx 2 dU dx = 0 potential energy is maximum = – ve viscous force, magnetic force etc., are non conservative forces. S.No. 1 (c) Neutral equilibrium : In the neutral equilibrium potential energy is constant. When a particle is displaced from its position it does not experience any force acting on it and continues to be in equilibrium in the displaced position. This is said to be neutral equilibrium. 2 3 Central in nature. Forces are velocitydependent and retarding in nature. 4 When only a conservative force acts within a systrem, the kinetic enrgy and potential energy can change. However their sum, the mechanical Work done against a non-conservative force may be dissipated as heat energy. 5 Work done is completely recoverable. Work done in not completely recoverable. Ex.13 The potential energy between two atoms in a a b molecule is given by, U(x) = 12 – 6 , where a and x x b are positive constants and x is the distance between the atoms. The system is in stable equilibrium when (A) x = 0 (C) x =  2a     b  a 2b (B) x= (D) x =  11a     5b  1/ 6 Non-Conservative forces Work done depends on Work done does not depend upon path path. Work done in round trip is Work done in a round zero. trip is not zero. Conservative forces Ex. 14 The figure shows three paths connecting points a and b. A single force F does the indicated work on a particle moving along each path in the indicated direction. On the basis of this information, is force F Sol. (C) Given that, U(x) = We, know a x F=– 12 – b x conservative? 6 du dx = (–12) a x–13 – (–6b) x–7 = 0 or or 6b x = 7 12a x13 6 x = 12a/6b = 2a/b 1/ 6 or x=  2a     b  Ans. No Explanation : NON-CONSERVATIVE FORCES For a conservative force, the work done in a round A force is said to be non-conservative if work done trip should be zero. by or against the force in moving a body depends upon the path between the initial and final positions. The frictional forces are non-conservative forces. Ex.15 The potential energy of a conservative system is This is because the work done against friction given by U = ax2 – bx where a and b are positive depends on the length of the path along which a body constants. Find the equilibrium position and discuss is moved. It does no depend only on the initial and whether the equilibrium is stable, unstable or neutral. final positions. Note that the work done by frictional force in a round trip is not zero. Work, Energy and Power Sol. In a conservative field F = – dU dx d Ex.17 A particle of mass 0.5 kg travels in a straight line with velocity v =ax3/2 where a = 5 m–1/2 s–1. What is the work done by the net force during its displacement from x = 0 to x = 2m?  F = – dx (ax2 – bx) = b – 2ax For equilibrium F = 0 or b – 2ax = 0  x  b 2a From the given equation we can see that d2U dx 2 = 2a (positive), i.e., U is minimum. Sol. m = 0.5 kg, v = ax3/2, a = 5 m–1/2 s–1, W = ? Initial velocity at x = 0, v0 = a × 0 = 0 is the stable equilibrium position. Final velocity at x = 2, v2 = a × 23/2 = 5 × 23/2 Ex.16 A force F = x2y2i + x2y2j (N) acts on a particle which moves in the XY plane. Work done = Increase in kinetic energy = Therefore, x = b 2a ( v 22 – v 02 ) = 1 2 1 2 m × 0.5 [(5 × 2 3/2)2 – 0] = 50 J. Ex.18 Figure shows two blocks A and B, each having a (a) Determine if F is conservative and mass of 320 g connected by a light string passing (b) find the work done by F as it moves the particle over a smooth light pulley. The horizontal surface on from A to C (fig.) along each of the paths ABC, which the block A can slide is smooth. The block A is ADC, and AC. attached to a spring of spring constant 40 N/m whose 5 5 a 2a other end is fixed to a support 40 cm above the [ Ans. (b) WABC = WADC = (J), WAC = (J) ] 5 3 horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find WORK-ENERGY THEOREM the velocity of the block A at the instant it breaks off According to work-energy theorem, the work the surface below it. Take g = 10 m/s2. done by all the forces on a particle is equal to the change in its kinetic energy. WC + WNC + WPS = K Where, W C is the work done by all the conservative forces. WNC is the work done by all non-conservative forces. WPS is the work done by all psuedo forces. Modified Form of Work-Energy Theorem We know that conservative forces are associated with the concept of potential energy, that is WC = U Sol. Let the block A start loosing contact with the surface So, Work-Energy theorem may be modified as below it at A´ after travelling a distance x as shown in WNC + WPS = K + U figure. WNC + WPS = E In this process the block B will shift from B to B’ such that BB’ = AA´ = x (as string is inextensible) and so there is a loss of gravitational potential energy = mgx. This energy is partly stored as elastic potential energy in the spring which is stretched by L and partly appears as kinetic energy of blocks A and B. So, by conservation of mechanical energy, we have mgx = or 1 1 mv2 + 2 2 v2 = gx – 2 mv + k 2m 1 2 k(L) Work, Energy and Power Ex.19 Figure shows a light, inextensible string attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the string is pulled over a pulley, of negligible mass and friction and fixed at height h = 3m from the ground level. The cart slides from x1 = 3 3 m to x2 = 4 m and during the move, tension in the string is kept constant 50 N. Find change in kinetic energy of the cart in joules. (Use 3 = 1.7) Ans. 50 Sol. Displacement of the point of 'A' of the string 2 = (3 3 ) 2  (3 )2 – 4 2  3 2 = 6 – 5 = 1 m (L)2 ...(i)  k = Work done by tension = 50 × 1 = 50 Joule. Now, for vertical equilibrium of block A at A’, So, substituting the value of L from Eq. (iii) in (ii) ENERGY DIAGRAMS Diagram, in which the total energy E and the potential energy U are plotted as functions of positions is called as energy diagram. The kinetic energy K = E – U is easily found by inspection. Since kinetic energy can never be negative, the motion of the system is constrained to regions where U < E. and solving for cos , we get (a) Energy Diagram for a harmonic oscillator. N + F cos  = mg But as for spring F = kL and for breaking off N = 0 the above equation reduces to kL cos  = mg cos  = 1 – mg kL ...(ii) =1– 0.32  10 40  0.40 = 4 5  L  0.4  5 So, that L =  cos   L  = – 0.4 = 0.1 m   4 and x = L tan  = 0.4 × 3 4 = 0.3 m Substituting these value of L and x in Equation (ii),  40  (0.1)2  10  0 . 3    v= 2  0.32   1/ 2 = 3  0.625 = 1.54 m/s U = Kx 2 2 Work, Energy and Power The potential energy of the block is U = Kx 2 is a parabola centered at the origin. Since, 2 the total energy is constant for a conservative system, E is represented by a horizontal straight line. Motion is limited to the shaded region where E > U; the limits of the motion x1 and x2 in the sketch, are sometimes called the turning points. The kinetic energy, K = E - U, is greatest at the origin. The particle flies past to a complete rest at one of the turning points x1, x2; then moves towards the origin with increasing kinetic energy, and the cycle is repeated. The harmonic oscillator provides a good example of bounded motion. As E increases, the turning point moves farther and farther away, but the particle can never move away freely. If E is decreased, the amplitude of motion decreases, until finally for E = 0 the particle lies at rest at x = 0. POWER Ans. Zero  When the time taken to complete a given amount of work is important, we measure the power of the agent doing work. The average power ( P or pav) delivered by an agent is given by P or pav = W t where W is the amount of work done in time t. Power is the ratio of two scalars- work and time. So, power is a scalar quantity. If time taken to complete a given amount of work is more, then power is less. For a short duration dt, if P is the power delivered during this duration, then   dS   = F =F. v dt  UNIT OF POWER A unit power is the power of an agent which does unit work in unit time. The power of an agent is said to be one watt if it does one joule of work in one second. 1 watt = 1 joule/section = 107 erg/second  This P is called as instantaneous power if dt is very small. 1 newton  1 metre 1 second [ Work ] [Power] = [Time] = = 1 N m s–1 . [ ML2 T 2 ] = [ML2 T–3] [T] Power has 1 dimension in mass, 2 dimensions in length and – 3 dimensions in time. S.No. 1 2 3 4 5 6 7 10  1 watt = Dimensional formula of power 9 where  is the smaller angle between F and v .   Power = F . v = Fv cos 90º = Zero. 8 By definition of dot product, P = Fv cos   Explanation. F and v are perpendicular.. Also, Power is defined as the time rate of doing work.   F  dS P dt Ex.20 A block moves in uniform circular motion because a cord tied to the block is anchored at the centre of a circle. Is the power of the force exerted on the block by the cord position, negative, or zero? 11 12 13 14 Human Activity Heart beat Sleeping Sitting Riding in a car –1 W alking (4.8 km h ) –1 C ycling (15 km h ) Playing Tennis Swimming (breaststroke, 1.6 km –1 h ) Skating Climbing Stairs (116 steps min –1 ) Cycling (21.3 km h –1 ) Playing Basketball Tube light Fan Pow er (W ) 1.2 83 120 140 265 410 440 475 535 685 700 800 40 60 Ex.21 What is represented by the slope of the worktime graph? Ans. Instantaneous power. Ex.22 What is represented by area under power-time graph? Ans. Work. Work, Energy and Power Sol. (a) Volume of wind flowing per second = Av Mass of wind flowing per second = Av Mass of air passing in t second = Avt 1 2 (b) Kinetic energy of air = mv2 = = (c) Electrical energy produced = Ex.23 What is the power of an engine which can lift 20 metric ton of coal per hour from a 20 metre deep mine? Sol. Mass, m = 20 metric ton = 20 × 1000 kg; Distance, S = 20 m; Time, t = 1 hour = 3600 s Power = Work Time = mg  S t = 20  1000  9.8  20 3600 watt = 1.09 × 103 W Electrical power 1 2 1 2 (Avvt)v2 Avv2t 25 100 × = Av 3 t 8 = Av 3t 8t 1 2 Av3t = Av 3  8 Now, A = 30 m 2 , v = 36 km h –1 = 36 × 5 18 m s –1 = 10 m s –1,  = 1.2 kg ms–1 Electrical power = 30  10  10  1.2 8 W = 4500 Ex.24 A one kilowatt motor pumps out water from a W = 4.5 kW well 10 metre deep. Calculate the quantity of water Ex.26 One coolie takes one minute to raise a box through pumped out per second. a height of 2 metre. Another one takes 30 second for the same job and does the same amount of work. 3 Sol. Power, P = 1 kilowatt = 10 watt Which one of the two has greater power and which S = 10 m ; Time, t = 1 second; Mass of water, m = ? one uses greater energy? Power = mg  S t  103 = m  9.8  10 1 = 3 or m = 10 9.8  10 Sol. P o w e r o f f i r s t c o o l i e = kg = 10.204 kg M  9 .8  2 60  M  9 .8  2    60   = M g S t M  9 .8  2 30 J s–1 = 2 J s–1 Power of second coolie Ex.25 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through in time t ? (b) What is the kinetic energy or the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km h–1 and the density of air is 1.2 kg m–3. What is the electrical power produced? Work Time = J s–1 = 2 × Power of first coolie So, the power of the second coolie is double that of the first. Both the coolies spend the same amount of energy. Aliter, We know that W = Pt P2 t1 For the same work, W = p1t1 = P2t2 or P = t 1 2 = 1 min ute 30 s = 2 or P2 = 2P1 Work, Energy and Power Ex.27 A large family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square metre. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8kW? Sol. If A m2 be the area, then power = 200 . A watts Useful electrical energy produced/s = 20 100 (200 A) = 40 . A watts But 40 A = 8000 or A = 200 m2 Ex.28 An elevator of total mass (elevator + passenger) 1800 kg is moving up with a constant speed of 2 ms–1. A frictional force of 4000 N oppose its motion. Determine the minimum power delivered by the motor to the elevator. Take g = 10 m s–2. Sol. Weight of (elevator + passenger) = mg = 1800 × 10 N = 18000 N Frictional force = 4000 N Total downward force on the elevator = (18000 + 4 0 00) N = 22000 N Clearly, the motor must have enough power to balance this force. Now, power, P = Fv = 2200 N × 2 m s-1 = 4400 W = 44000 746 hp = 58.98 hp