Work, Energy, Power and Collision PDF

E3 60 32 Work, Energy, Power and Collision 6.1 Introduction. The terms 'work', 'energy' and 'power' are frequently used in everyday language. A farmer clearing weeds in his field is said to be working hard. A woman carrying water from a well to ID her house is said to be working. In a drought affect...

E3 60 32 Work, Energy, Power and Collision 6.1 Introduction. The terms 'work', 'energy' and 'power' are frequently used in everyday language. A farmer clearing weeds in his field is said to be working hard. A woman carrying water from a well to ID her house is said to be working. In a drought affected region she may be required to carry it over large distances. If she can do so, she is said to have a large stamina or energy. Energy is thus the capacity to do work. The term power is usually associated with speed. In karate, a U powerful punch is one delivered at great speed. In physics we shall define these terms very precisely. We shall find that there is at best a loose correlation between the physical definitions D YG and the physiological pictures these terms generate in our minds. Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force. 6.2 Work Done by a Constant Force. Let a constant force F be applied on the body such that it makes an angle  with the U horizontal and body is displaced through a distance s By resolving force F into two components : ST (i) F cos in the direction of displacement of the body. (ii) F sin in the perpendicular direction of displacement of the body. F sin F  F cos s Since body is being displaced in the direction of F cos  , therefore work done by the force in displacing the body through a distance s is given by W  (F cos  ) s  Fs cos  or W  F.s Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body. Work, Energy Power, and Collision 33 If a number of force F 1 , F 2 , F 3...... F n are acting on a body and it shifts from position vector r 1 to position vector r 2 then W  (F 1  F 2  F 3 .... F n ).( r 2  r 1 ) 6.3 Nature of Work Done. Negative work its Negative work means that force (or component) is opposite to displacement i.e. Direction of motion F F Direction of motion  s 0 o    90 o its s 90 o    180 o ID  (or E3 Positive work means that force component) is parallel to displacement 60 Positive work The negative work signifies that the external force opposes the motion of the body. Example: (i) When a person lifts a body from the Example: (i) When a person lifts a body from the ground, the work done by the (upward) lifting force is positive ground, the work done by the (downward) force of gravity is negative. D YG F man U The positive work signifies that the external force favours the motion of the body. Fg s U (ii) When a lawn roller is pulled by applying a force along the handle at an acute angle, work done by the applied force is positive. s (ii) When a body is made to slide over a rough surface, the work done by the frictional force is negative. F F s ST s (iii) When a spring is stretched, work done by the external (stretching) force is positive. s F Maximum work : Wmax  F s (iii) When a positive charge is moved towards another positive charge. The work done by F electrostatic force between them is negative. + + s Minimum work : Wmin   F s 34 Work, Energy, Power and Collision When cos   maximum  1 i.e.   0 o When cos   minimum  1 i.e   180 It means force does maximum work when angle between force and displacement is zero. It means force does minimum o [maximum Zero work E3 Under three condition, work done becomes zero W  Fs cos   0 60 negative] work when angle between force and displacement is 180 o. (1) If the force is perpendicular to the displacement [ F  s ] ID Example: (i) When a coolie travels on a horizontal platform with a load on his head, work done against gravity by the coolie is zero. s Fg (ii) When a body moves in a circle the work done by the centripetal force is always zero. (iii) In case of motion of a charged particle in a U magnetic field as force [F  q(v  B)] is always perpendicular to motion, work done by this force is always zero. D YG (2) If there is no displacement [s = 0] F Example: (i) When a person tries to displace a wall or heavy stone by applying a force then it does not move, the work done is zero. s 0 (ii) A weight lifter does work in lifting the weight off the ground but does not work in holding it up. (3) If there is no force acting on the body [F = 0] U Example: Motion of an isolated body in free space. ST Sample Problems based on work done by constant force Problem 1. A body of mass 5 kg is placed at the origin, and can move only on the x-axis. A force of 10 N is acting on it in a direction making an angle of 60 o with the x-axis and displaces it along the x-axis by 4 metres. The work done by the force is (a) 2.5 J (b) 7.25 J (c) 40 J  Solution : (d) Work done  F.s  F s cos   10  4  cos 60 o  20 J Problem 2. (d) 20 J A force F  (5ˆi  3 ˆj) N is applied over a particle which displaces it from its origin to the point r  (2ˆi  1ˆj) metres. The work done on the particle is (a) –7 J (b) +13 J (c) +7 J  Solution : (c) Work done  F.r  (5i  3ˆj).(2ˆi  ˆj)  10  3  7 J (d) +11 J Work, Energy Power, and Collision 35 Problem 3. A horizontal force of 5 N is required to maintain a velocity of 2 m/s for a block of 10 kg mass sliding over a rough surface. The work done by this force in one minute is (a) 600 J (b) 60 J (c) 6 J (d) 6000 J Solution : (a) Work done = Force  displacement = F  s = F  v  t = 5  2  60 = 600 J. Problem 4. (a) 28 J (b) 8 J (c) 18 J Solution : (a) Work done to displace it horizontally = F  s = 8  1 = 8 J 60 A box of mass 1 kg is pulled on a horizontal plane of length 1 m by a force of 8 N then it is raised vertically to a height of 2m, the net work done is (d) None of above  Net work done = 8 +20 = 28 J Problem 5. E3 Work done to raise it vertically F  s = mgh = 1  10  2 = 20 J A 10 kg satellite completes one revolution around the earth at a height of 100 km in 108 minutes. The work done by the gravitational force of earth will be (b) 108  10 J 100 100  10 J 108 ID (a) 108  100  10 J (c) (d) Zero Solution : (d) Work done by centripetal force in circular motion is always equal to zero. U 6.4 Work Done by a Variable Force. When the magnitude and direction of a force varies with position, the work done by such a D YG force for an infinitesimal displacement is given by dW  F. d s The total work done in going from A to B as shown in the figure is W  B A F. d s   B (F cos  )ds A B In terms of rectangular component F  Fx ˆi  Fy ˆj  Fz kˆ ds  U d s  dx ˆi  dyˆj  dz kˆ  (F ˆi  F ˆj  F kˆ ).(dxˆi  dyˆj  dzkˆ ) or W   x A xB y Fx dx  z  yB ST Problem 6. A B W xA F yA Fy dy   zB zA Fz dz Sample Problems based on work done by variable force  A position dependent force F  (7  2 x  3 x 2 ) N acts on a small abject of mass 2 kg to displace it from x  0 to x  5m. The work done in joule is (a) 70 J Solution : (d) Work done  Problem 7. (b) 270 J  x2 x1 F dx   5 0 (c) 35 J (d) 135 J (7  2 x  3 x 2 )dx  [7 x  x 2  x 3 ]50  35  25  125  135 J A particle moves under the effect of a force F = Cx from x = 0 to x = x1. The work done in the process is 36 Work, Energy, Power and Collision [CPMT 1982] (b) Solution : (b) Work done   x2 F dx  x1 Problem 8.  x1 0 1 Cx 12 2 (c) Cx 1 (d) Zero x x2  1 1 Cx dx  C    C x 12 2 2   0 60 (a) Cx 12 The vessels A and B of equal volume and weight are immersed in water to a depth h. The vessel A has an opening at the bottom through which water can enter. If the work done in immersing A and B are W A and W B respectively, then (b) W A  WB (c) W A  WB (d) E3 (a) W A  WB WA    WB Solution : (b) When the vessels are immersed in water, work has to be done against up-thrust force but due to opening at the bottom in vessel A, up-thrust force goes on decreasing. So work done will be less in this case. Work done in time t on a body of mass m which is accelerated from rest to a speed v in time t1 as a function of time t is given by 1 v 2 m t 2 t1 (b) m v 2 t t1 (c) 1  mv  2  t1 U (a) ID Problem 9. 2  2  t  D YG 1 v 1 1  Solution : (d) Work done = F.s = ma. a t 2   m a 2 t 2  m  2 2 2  t1   2  2  t  (d) 1 v2 2 m t 2 t12   v given   As accelerati on (a)  t 1   6.5 Dimension and Units of Work. Dimension : As work = Force  displacement  [W] = [Force]  [Displacement]  [MLT 2 ]  [L]  [ML2 T 2 ] Units : The units of work are of two types U Absolute units ST Joule [S.I.]: Work done is said to be one Joule, when 1 Newton force displaces the body through 1 meter in its own direction. From W = F.s 1 Joule = 1 Newton  1 metre Erg [C.G.S.] : Work done is said to be one erg when 1 dyne force displaces the body through 1 cm in its own direction. From W = F s 1 Erg  1Dyne  1cm Relation between Joule and erg 1 Joule = 1 N  1 m = 105 dyne  102 cm = 107 dyne  cm = 107 Erg Gravitational units kg-m [S.I.]: 1 Kg-m of work is done when a force of 1kg-wt. displaces the body through 1m in its own direction. From W=Fs 1 kg-m = 1 kg-wt  1 metre = 9.81 N  1 metre = 9.81 Joule gm-cm [C.G.S.] : 1 gm-cm of work is done when a force of 1gm-wt displaces the body through 1cm in its own direction. From W = F s 1 gm-cm = 1gm-wt  1cm. = 981 dyne  1cm = 981 erg Work, Energy Power, and Collision 37 6.6 Work Done Calculation by Force Displacement Graph. Let a body, whose initial position is x i , is acted upon by a variable force (whose magnitude is changing continuously) and consequently the body acquires its final position x f. 60 Let F be the average value of variable force within the interval dx from position x to (x + dx) i.e. for small displacement dx. The work done will be the area of the shaded strip of width dx. The work done on the body in displacing it from position x i to x f will be equal to the sum of Force areas of all the such strips  F W   xf  xf xi xi dW   xf xi F dx O (Area of strip of width dx ) W  Area under curve Between x i and x f xi x xf Displacemen t dx ID W  E3 dW  F dx i.e. Area under force displacement curve with proper algebraic sign represents work done by the force. U Sample problems based on force displacement graph Problem 10. A 10 kg mass moves along x-axis. Its acceleration as a function of its position is shown in (a) 8  10 2 J (b) 16  10 2 J 20 15 10 5 a (cm/sec2) D YG the figure. What is the total work done on the mass by the force as the mass moves from x  0 to x  8 cm [AMU (Med.) 2000] U (c) 4  10 4 J (d) 0 2 4 6 8 x (cm) 1.6  10 3 J ST Solution : (a) Work done on the mass = mass  covered area between the graph and displacement axis on a-t graph. = 10  1 (8  10  2 )  20  10  2 = 8  10 2 J. 2 Problem 11. The relationship between force and position is shown in the figure given (in one (a) 20 ergs (b) 60 ergs Force (dyne) dimensional case). The work done by the force in displacing a body from x  1 cm to [CPMT 1976] x  5 cm is 20 10 0 10 20 1 2 3 4 5 6 x (cm) 38 Work, Energy, Power and Collision (c) 70 ergs (d) 700 ergs Solution : (a) Work done = Covered area on force-displacement graph = 1  10 + 1  20 – 1  20 + 1  10 = 20 erg. 60 Problem 12. The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. The mass of the body is 25 kg and initial velocity is 2 m/s. When the distance covered by the body is 5m , its kinetic energy would be F (Newton) 20 E3 (a) 50 J 10 (b) 40 J 0 (d) 10 J Solution : (d) Initial kinetic energy of the body  1 2 3 4 x (m) ID (c) 20 J 1 1 mu 2   25  (2) 2  50 J 2 2 1  4  20  50  40  10 J. 2 D YG  50  U Final kinetic energy = Initial energy – work done against resistive force (Area between graph and displacement axis) 6.7 Work Done in Conservative and Non-Conservative Field. (1) In conservative field work done by the force (line integral of the force i.e.  F.dl ) is independent of the path followed between any two points. U W AB  W AB  W AB B Path III II  F.d l   F.d l   F.d l III Path I Path II ST or I A Path I Path II Path III (2) In conservative field work done by the force (line integral of the force i.e.  F.dl ) over a closed path/loop is zero. W A B  WB A  0 or A B  F.dl  0 Conservative force : The forces of these type of fields are known as conservative forces. Work, Energy Power, and Collision 39 Example : Electrostatic forces, gravitational forces, elastic forces, magnetic forces etc and all the central forces are conservative in nature. If a body of man m lifted to height h from the ground level by different path as shown in B B II I B B III l IV A A A WI  F. s  mg  h  mgh A ID Work done through different paths h E3  60 the figure WII  F. s  mg sin   l  mg sin   h  mgh sin   D YG WIV  F. d s  mgh U WIII  mgh 1  0  mgh 2  0  mgh 3  0  mgh 4  mg (h1  h 2  h3  h4 )  mgh It is clear that WI  WII  WIII  WIV  mgh. Further if the body is brought back to its initial position A, similar amount of work (energy) is released from the system it means W AB  mgh and WBA  mgh. Hence the net work done against gravity over a round strip is zero. W Net  W AB  WBA U  mgh  (mgh )  0 i.e. the gravitational force is conservative in nature. ST Non-conservative forces : A force is said to be non-conservative if work done by or against the force in moving a body from one position to another, depends on the path followed between these two positions and for complete cycle this work done can never be a zero. Example: Frictional force, Viscous force, Airdrag etc. If a body is moved from position A to another position B on a rough table, work done against frictional force shall depends on the length of the path between A and B and not only on the position A and B. W AB  mgs Further if the body is brought back to its initial position A, work has to be done against the R s F 40 Work, Energy, Power and Collision frictional force, which always opposes the motion. Hence the net work done against the friction over a round trip is not zero. WBA  mgs. W Net  W AB  WBA  mgs  mgs  2 mgs  0. 60 i.e. the friction is a non-conservative force. Sample problems based on work done in conservative and non-conservative field Problem 13. If W1 , W2 and W3 represent the work done in moving a particle from A to B along three E3 different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation (a) W1  W2  W3 ID m (b) W1  W2  W3 (c) W1  W2  W3 1 2 3 A W2  W1  W3 U (d) B D YG Solution : (b) As gravitational field is conservative in nature. So work done in moving a particle from A to B does not depends upon the path followed by the body. It always remains same. Problem 14. A particle of mass 0.01 kg travels along a curve with velocity given by 4ˆi  16 kˆ ms-1. After some time, its velocity becomes 8ˆi  20 ˆj ms 1 due to the action of a conservative force. The work done on particle during this interval of time is (a) 0.32 J (b) 6.9 J (c) 9.6 J (d) 0.96 J Solution : (d) v1  4 2  16 2  272 and v 2  8 2  20 2  464 U Work done = Increase in kinetic energy  1 1 m[v 22  v 12 ]   0.01[464  272 ]  0.96 J. 2 2 ST 6.8 Work Depends on Frame of Reference. With change of frame of reference (inertial) force does not change while displacement may change. So the work done by a force will be different in different frames. Examples : (1) If a porter with a suitcase on his head moves up a staircase, work done by the upward lifting force relative to him will be zero (as displacement relative to him is zero) while relative to a person on the ground will be mgh. (2) If a person is pushing a box inside a moving train, the work done in the frame of train will F.s while in the frame of earth will be F. (s  s 0 ) where s 0 is the displacement of the train relative to the ground. h Work, Energy Power, and Collision 41 6.9 Energy. The energy of a body is defined as its capacity for doing work. (2) Dimension: [ML2 T 2 ] it is same as that of work or torque. (3) Units : Joule [S.I.], erg [C.G.S.] 60 (1) Since energy of a body is the total quantity of work done therefore it is a scalar quantity. Practical units : electron volt (eV), Kilowatt hour (KWh), Calories (Cal) 1 eV = 1.6  10 19 Joule 1 KWh = 3.6  10 6 Joule 1 Calorie = 4. 18 Joule E3 Relation between different units: 1 Joule = 10 7 erg ID (4) Mass energy equivalence : Einstein’s special theory of relativity shows that material particle itself is a form of energy. The relation between the mass of a particle m and its equivalent energy is given as where c = velocity of light in vacuum. U E  mc 2 If m  1 amu  1.67  10 27 kg then E  931 MeV  1.5  10 10 Joule. D YG If m  1kg then E  9  10 16 Joule Examples : (i) Annihilation of matter when an electron (e  ) and a positron (e  ) combine with each other, they annihilate or destroy each other. The masses of electron and positron are converted into energy. This energy is released in the form of  -rays. e  e   Each  photon has energy = 0.51 MeV. U Here two  photons are emitted instead of one  photon to conserve the linear momentum. (ii) Pair production : This process is the reverse of annihilation of matter. In this case, a photon ( ) having energy equal to 1.02 MeV interacts with a nucleus and give rise to electron ST (e  ) and positron (e  ). This energy is converted into matter.  (Photon) e– + e + (iii) Nuclear bomb : When the nucleus is split up due to mass defect (The difference in the mass of nucleons and the nucleus) energy is released in the form of  -radiations and heat. (5) Various forms of energy (i) Mechanical energy (Kinetic and Potential) (ii) Chemical energy (iii) Electrical energy (iv) Magnetic energy (v) Nuclear energy (vi) Sound energy 42 Work, Energy, Power and Collision (vii) Light energy (viii) Heat energy (6) Transformation of energy : Conversion of energy from one form to another is possible through various devices and processes. – Photoelectric cell Primary cell Sounds Electrical Heat electrical U ID Chemical heat Coal Burning Microphone Motor ST U Electrical Sound Speaker Fe Hot Voltameter G Cu Electrical Heat Heater Electrical Chemical Anod e+ Cold Thermo-couple Electrical Mechanical D YG Heat Mechanical Cathod + e Anode S Dynamo Engine Chemical electrical E3 N Light Electrical 60 Mechanical electrical Electrical Light Catho – de Electrolyt e Bulb Sample problems based on energy Problem 15. A particle of mass ‘m’ and charge ‘q’ is accelerated through a potential difference of ‘V’ volt. Its energy is [UPSEAT 2001] Work, Energy Power, and Collision 43 (a) qV q (c)  V m  (b) mqV q mV (d) Solution : (a) Energy of charged particle = charge  potential difference = qV Problem 16. An ice cream has a marked value of 700 kcal. How many kilowatt hour of energy will it deliver to the body as it is digested (b) 0.90 kWh Solution : (a) 700 k cal = 700  10 3  4.2 J  700  10 3  4.2 3.6  10 6 (c) 1.11 kWh  0.81 kWh (d) 0.71 kWh 60 (a) 0.81 kWh [As 3.6  10 6 J  1 kWh ] E3 Problem 17. A metallic wire of length L metres extends by l metres when stretched by suspending a weight Mg to it. The mechanical energy stored in the wire is (a) 2 Mgl (b) Mgl (c) Mgl 4 (d) 1 Mgl. Fx  2 2 ID Solution : (c) Elastic potential energy stored in wire U  Mgl 2 6.10 Kinetic Energy. The energy possessed by a body by virtue of its motion is called kinetic energy. U Examples : (i) Flowing water possesses kinetic energy which is used to run the water mills. (ii) Moving vehicle possesses kinetic energy. D YG (iii) Moving air (i.e. wind) possesses kinetic energy which is used to run wind mills. (iv) The hammer possesses kinetic energy which is used to drive the nails in wood. (v) A bullet fired from the gun has kinetic energy and due to this energy the bullet penetrates into a target. (1) Expression for kinetic energy : Let m = mass of the body, u = Initial velocity of the body (= 0) F = Force acting on the body, a = Acceleration of the body U s = Distance travelled by the body, v = Final velocity of the body F v u= 0 ST From v 2  u 2  2as  v 2  0  2as s  s v2 2a Since the displacement of the body is in the direction of the applied force, then work done by the force is W  F  s  ma   W v2 2a 1 mv 2 2 This work done appears as the kinetic energy of the body KE  W  1 mv 2 2 44 Work, Energy, Power and Collision (2) Calculus method : Let a body is initially at rest and force F is applied on the body to displace it through d s along its own direction then small work done dW  F.d s  F ds dW  m a ds  dW  m  dv    As a  dt    dv ds dt dW  mdv. ds dt dW  m v dv …….(i) E3  [As F = ma] 60  ds    As dt  v    Therefore work done on the body in order to increase its velocity from zero to v is given by v v 2  1 W  mv dv  m v dv  m    mv 2 0 0 2  2 0  v ID  v 1 m (v. v ) 2 D YG In vector form KE  U This work done appears as the kinetic energy of the body KE  1 mv 2. 2 As m and v.v are always positive, kinetic energy is always positive scalar i.e. kinetic energy can never be negative. (3) Kinetic energy depends on frame of reference : The kinetic energy of a person of mass 1 m, sitting in a train moving with speed v, is zero in the frame of train but mv 2 in the frame of 2 the earth. U (4) Kinetic energy according to relativity : As we know E  1 mv 2. 2 But this formula is valid only for (v