1. By Jenoy MathewA.K.M.H.S.S,Kottoor

2. Work The work done by a force is measured by the dot product of force and displacement. If a force F acting on a body displaces it through a displacement d,then the work done by the force is given by W=F.d 3. F ddWork done(W)=F.d 4. FsinFFcosdWork done(W)=Fcos.d 5. Work is a scalar The body is not displaced in the perpendicular direction. So the work done by Fsin component of force is zero. Work done is the dot product of two vectors. So work done is a scalar quantity. 6. Types of Work 7. F Angled F.d=Fdcos between Fand d is zero=Fd cosO=FdWork done is +ve 8. W=F.d=F.d cos =F.d cos 900=zero F =90o dWork done is Zero 9. W= F.dcos =1800 =F.dcos1800= -Fd FdThe work done is -ve 10. Kinetic Energy 11. Kinetic EnergyIt is the energy possessed by virtue of its motion 12. Regarding with K.E. K.E. is always a +ve scalar quantity. K.E. depends on the frame of reference. K.E.related to its momentum as K.E.=p2/2m,where m is the mass and p is the momentum 13. Work energy principle Work done by a force in displacing a body measures the change in kinetic energy of the body 14. U=O V m S By equation ofmotion, v2= u2+2aSU=O , So v2= 2aS2W = F.S = ma X v /2a S = v2/2a=1/2mv2 15. This work done appears as the K.E. Of thebody, i.e.,K.E.=1/2mv 2 16. Work done by a variable force 17. BWork done for theForcedisplacementdx, dw=F.dxADCXfO Xi dxDisplacement The total work done on the body from A to B, W= = Area ABCD 18. Work energy theorem for avariable force 19. The time rate of change of K.E. is, dK/dt=d/dt(1/2 mv2)=1/2m(2v).dv/dt=m dv/dt.v=F.v=F.dx/dt = i.e,Ki Kf ==WdK= F.dx 20. Problems 21. Calculate the work done in lifting a mass of5kg vertically through 8m Force on the mass = weight of the body = 5 X 9.8 N Vertical lift = 8m Work done = Force X vertical lift = 5 X 9.8 X 8 = 392 J 22. A man pushes a roller with a force of 50 N througha distance of 20m.Calculate the work done if thehandle of the roller is inclined at an angle of 60owith the ground? W = F X S cos=50 X 20X cos60 = 500J 23. Calculate the work done in raising a stone of mass 5kgand specific gravity 3 lying at the bed of a lake througha height of 5m? Loss of weight of stone in water = =kg wt. weight of stone in water = kg wt = kg wt Force,F = kg wt = X 9.8 NW= X 9.8 X 5 = 163.3 J 24. A bullet weighing 10g is fired with a velocity of800m/s.After passing through a mud wall 1m thick, itsvelocity decreases to 100m/s.Find the averageresistance offered by the mud wall? W = 1/2mv2 1/2mu2 F.d = 1/2mv2 1/2mu2F= { v2 - u2}= -3150N 25. Potential energy 26. Potential energy The energy possessed by a body by virtue of its position (or configuration) in a field or state of strain is called Potential energy. e.g. Water stored in the reservoir of a dam, an apple hanging on an apple tree. 27. Gravitational potential energyThe energy possessed by a body dueits position in the gravitational fieldof earth is called gravitationalpotential energy. 28. h180o W=F.hmg = F h cos = mgh cos 1800 h= - mghP.E. = V = mgh Earth 29. Potential energy of a springWhen a spring is stretched orcompressed from its normalposition(x = 0),a restoring force is set up inside the spring 30. The same spring is stretched orcompressed as shown below. In which case does the force exerted by the spring havethe largest magnitude? 31. This restoring force is directly proportional tothe displacement from the equilibriumposition. i.e., F - x F = -kx,where k is a constant called springconstant. ve sign indicates that restoring force is opposite to the displacement. 32. The external force required to keep the spring at a displacement x is Fext = -F = +kx Therefore the total work done to stretch the spring from x = 0 to x = x is, W = = = 1/2 kx2 33. From the graphForcekx x Displacement 34. For a particular displacement x,the area of the graph = area of OAB area of OAB = 1/2 x kx = 1/2 kx2 = P.E. 35. Conservative forcesand Non-conservative forces 36. Conservative forces if the amount of work done against aforce depends only on the initial andfinal positions of the body moved andnot the path followed,then such a forceis called conservative force. e.g. gravitational force,electrostaticforce,elastic force,Magneticforce,Lorentz force 37. Non-conservative If the workdone against a force depends on the path followed,such a force is called non-conservative force. e.g. Frictional force,Viscous force,Damping force,A force that generally depends on velocity 38. Conservation of mechanical energy Energy can neither be created nor destroyed.It can only be transformed from one form into another without gain or loss.i.e. the total energy of a closed system remains constant. 39. At A, A K.E.=1/2 mu2 = 0P.E.(V) = mghTotal energy= K.E. +P.E. = mgh xAt B, P.E. = mg(h x) K.E. = mv2hFor a free falling body,v =B2gxSo,K.E. = m X 2gx = mgx Total energy = K.E.+P.E. = mg (h- x) + mgx C = mgh 40. At C, P.E. = O Energy Here velocity at C(V)=2ghSo K.E. = mghTotal energy = mgh+ O = mghT.E. K.E Variation of K.E.,P.E.P.E. And total energy with heightHeight 41. Problems 42. A ball is thrown vertically downward from a height of 20m with aninitial velocity u.After collision with the ground, it losses 40% of itsinitial velocity and rebounds to the same height. Find out u ? Total energy of the ball when just throwndownwards = 1/2mu2 + mgh. Energy after collision = (1/2mu2 + mgh)60/10o Energy of the ball at a height h after collision = mgh By conservation ofenergy, (1/2mu2 + mgh)60/10o =mgh 3/2mu2 = 2 mgh, i.e., u2 = 4/3 gh u = 4/3 9.8 20 = 16.17 m/s 43. A body of mass 1kg initially at rest is droppedfrom a height of 2m on to a vertical spring havingforce constant 490 N/m.Calculate the maximumdistance through which the spring will becompressedLoss in gravitational P.E. Of the body = Gain in the P.E. Of the spring Mg (h + x) = 1/2kx21 9.8(2+x) = 490 x2100 x2 4x 8 = 0,on solving x = 0.303m 44. A rain drop of radius 2mm falls from a height of 250m above theground.What is the work done by the gravitational force on thedrop ? Volume,V = 4/3 r3 Mass, m = V d Work done by gravitational force(W) = mgh= 4/3 r3dg h W = 4/3 (2 10-3 )3 103 9.8 250 = 0.082 J 45. Different forms of energy 46. Light energy When an excited electron in an atom or molecule makes a transition to a lower energy level,this form of energy is produced Light transmitted as photons of energy E = h 47. Internal energy The molecules of a body are in randommotion.The molecules thus possess K.E. The intermolecular attraction produces P.E. The sum of these kinetic and potentialenergies of the system is called internalenergy. 48. Heat or thermal energy Heat is a form of energy An object possesses heat energy due to the molecules moving in it 49. Nuclear energy The binding energy ofnucleons(neutrons and protons) in thenucleus is called nuclear energy During nuclear reactions like nuclearfusion or fission, this energy is released 50. Electrical energy Electric charges develop an attraction orrepulsion among them A work is to be done to move a charge inan electric field This work is stored as electrical energy 51. Mass energy equivalence When a mass m is converted into energy E The corresponding energy released is E = mc2, Where c is the velocity of light This equation is known as Einsteins mass energyequivalence If 1 kg of matter is converted as energy,then E = 1 ( 3 108 )2 = 9 1016 J The energy released during nuclear reaction isdue to the conversion of mass into energy 52. About 4 1010 kg matter per second is convertedinto energy in the sun.What is the poweroutput of the sun? E = mc2 = 4 1010 ( 3 108 )2 = 3.6 1027 J This much of energy is liberated per second Power out put of sun = 3.6 1027 J/s = 3.6 1027 W 53. Power Power is defined as the rate at which work is done If W is the work in a time t,the average power,P= W/t If dw is the work done in a small time interval dt,then the instantaneous power, P= dw/dt Power is a scalar quantity P=W/t = F.S/t= F.v (since v=S/t) 54. Problems 55. A water pump driven by petrol raises water at a rate of0.5 m3/min from a depth of 30m.If the pump is 70%efficient,what power is developed by the engine Volume of water taken/s = 0.5/60 m3/s Mass of water taken/s = 0.5/60 X 103 kg/s Out put power=mgh/t=0.5/60 X 103 X9.8 X 30=2450W i.e , Input power = output power/efficiency 2450/70/100=2450 X 100 /70 = 3500 W 56. A railway engine of mass 12000kg is moving at aconstant speed 5m/s up an inclined plane of150.Calculate the power of the engine?(Take g=9.8 m/s) RF Mg sinMg cos Mg 150 57. Force (F) = mg sin F = 12000 X 9.8 X sin 150 = 3.04 X 104 N P = F X v = 3.04 X 104 N X v P = F X v = 3.04 X 104 X 5 = 152000 W = 152 kW 58. Collisions 59. The abrupt change to the path of amoving body(or bodies)due to itsinteraction with other body(or bodies)iscalled collision. The magnitude and direction of thevelocity of the colliding bodies maychange in a collision. 60. Types of collisions Perfectly elastic collision The collision in which,both K.E. and linear momentum of the colliding bodies remain conserved is called perfectly elastic collision. 61. Inelastic collision The collision in which linear momentum of the system remains conserved but its kinetic energy is not conserved is called inelastic collision. 62. Perfectly inelastic collisions A collision is said to be perfectly inelastic if after collision,the two bodies stick together an move as a single system 63. Elastic collision in One Dimensionm1m2 u1u2 At the time ofcollisionBefore collision 64. After collision V1 V2m1m2The collision is elastic, the K.E. And linear momentum remain conserved 65. By conservation of momentum,m1 u 1+m2 u2 = m1 v 1+m2 v2 m1(u 1 - v 1 ) = m2 (v2 - u2) ------- (1)By conservation of K.E.,1/2 m1 u 1 2+1/2m2 u2 2 = 1/2m1 v 1 2+1/2 m2 v2 2 m1(u 12 - v 12 ) = m2 (v22 -u22)m1(u 1 + v 1 ) (u 1 - v 1 ) = m2 (v2 + u2) m2 (v2 - u2)-----(2) Dividing eq (2) by (1), u 1 + v 1 = v2 + u2 -------(3) i.e., u 1 - u2 = v2 - v 1 -------(4) 66. (u 1 - u2 ) is the relative velocity of approach and (v2 - v 1 ) is the relative velocity ofseparation 67. Velocities after collisionFrom equation ( 3) ,v2 = u 1 + v 1 - u2Substituting the value of v2 in Eq.(1), m1(u 1 - v 1 ) = m2 (u 1 + v 1 - u2- u2)m1 u 1 - m1 v 1= m2 u 1 +m2 v 1- 2m2 u2m1 u 1- m2 u 1+ 2m2 u2 = v 1( m1+ m2 ) 68. There fore,v 1 ={ ( m1- m2 ) u 1 / ( m1+ m2 ) }+ {2 m2 u2/ ( m1+ m2 )} Similarly we can see that,v 2 ={ ( m2- m1 ) u 2 / ( m1+ m2 ) }+ {2 m1 u1/ ( m1+ m2 )} 69. Special cases When the two bodies have equal masses,i.e., m1 = m2 = mv 1 = 2m u2/2m = u2 and v 2 = u 1 . So if the two bodieshave equal masses after elastic collision in onedimension,exchange their velocities If the second body is at rest before collision.then, u2 =O, also m1 = m2 = m ,after collision v 1 = O and v 2 = u 1 70. If m1>> m2 ,we can neglect m2 and u2 = O,then v 1 = m1 u 1 / m1 = u 1 , v 2 = 2m1 u 1 / m1 = 2 u 1 .i.e.,If a heavy body collides with a light body,there is no change in the velocity for the heavy body and the light body moves with twice the velocity of the heavy body 71. If m1