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  1. 1. 1 Chapter No. 05 Mohd Yousuf Soomro ( Lecturer ) Applications Of Newton’s Laws Institute of Physics 5-1 Force laws: To understand the effects of forces a detailed microscopic understanding of interaction of objects with their environment is necessary. All of known forces in the universe can be grouped into four types. 1- The Gravitational forces 2- The Electromagnetic forces 3- The Weak Nuclear forces 4- The Strong Nuclear forces. 1- The Gravitational forces: It originates with the presence of matter. The gravitational force between two protons just touching their surfaces is about 10-38 of the strong force between them. The principal difference between gravitation & other forces is that, on the practical scale, gravity is cumulative and infinite in range. Gravity is the weakest of the four fundamental forces, yet it is the dominant force in the universe for shaping the large scale structure of galaxies, stars, etc. The gravitational force between two masses m1 and m2 is given by the relationship: This is often called the "universal law of gravitation" and G the universal gravitation constant. It is an example of an inverse square law force. The force is always attractive and acts along the line joining the centers of mass of the two masses. The forces on the two masses are equal in size but opposite in direction, obeying Newton's third law. Viewed as an exchange force, the mass less exchange particle is called the graviton. 1
  2. 2. 2 The gravity force has the same form as Coulomb's law for the forces between electric charges, i.e., it is an inverse square law force which depends upon the product of the two interacting sources. 2- The Electromagnetic forces: Electromagnetism is important in the structure & the interaction of fundamental particle. It is also responsible for the binding of atoms & the structure of solids. the electromagnetic force manifests itself through the forces between charges. The electromagnetism force between two neighboring protons is about 10-2 of the strong force. The electromagnetic force is a force of infinite range which obeys the inverse square law, and is of the same form as the gravity force. 3- The Weak Nuclear forces: The weak force is responsible for certain radioactive process & other similar decay processes involving fundamental particles. The weak force between two neighboring protons is about 10-7 of strong force between them. One of the four fundamental forces, the weak interaction involves the exchange of the intermediate vector bosons, the W and the Z. Since the mass of these particles is on the order of 80 GeV, the uncertainty principle dictates a range of about 10-18 meters which is about 0.1% of the diameter of a proton. It is crucial to the structure of the universe in that 1. The sun would not burn without it since the weak interaction causes the transmutation p ≥ n so that deuterium can form and deuterium fusion can take place. 2. It is necessary for the buildup of heavy nuclei. 2
  3. 3. 3 It was in radioactive decay such as beta decay that the existence of the weak interaction was first revealed. 4- The Strong Nuclear force: A force which can hold a nucleus together against the enormous forces of repulsion of the protons is strong indeed. However, it is not an inverse square force like the electromagnetic force and it has a very short range. It is responsible for the binding of the nuclei, is the dominant force in the reaction & decays of most of the fundamental particles. The relative strength b/w two neighboring protons is 1. the strong force is 1038 times greater than gravitational force. Fundamental Forces Unification for forces: In 19th century James Clerk Maxwell united the separate electric & magnetic fields into a single electromagnetism force. 3
  4. 4. 4 In 1967 a theory was proposed by “Dr. Abdus Salam”, “Stephen Weinberg”. According to this theory the weak & electromagnetism forces could be regarded a part of a single force called electroweak force. Theories called GUTs (Grand Unification Theories) predicts about the unification of strong and electroweak forces. Other theory called TOE ( Theory of everything ) which unified the Strong, electroweak & gravitational forces. Prediction of TOE: One prediction of these theories is that proton should not be a stable particle but should decay on a time scale greater than 1031 years. One way to test this theory is to which collection of 5.3 X 10 33 protons from a year to if one of proton decays. Electroweak Unification The discovery of the W and Z particles, the intermediate vector bosons, in 1983 brought experimental verification of particles whose prediction had already contributed to the Nobel prize awarded to Weinberg, Salam, and Glashow in 1979. The photon , the particle involved in the electromagnetic interaction, along with the W and Z provide the necessary pieces to unify the weak and electromagnetic interactions. With masses around 80 and 90 GeV, respectively, the W and Z were the most massive particles seen at the time of discovery while the photon is mass less. The discovery of the W and Z particles in 1983 was hailed as a confirmation of the theories which connect the weak force to the electromagnetic force in electroweak unification Grand Unification Grand unification refers to unifying the strong interaction with the unified electroweak interaction. One prediction of the grand unified theories is that the proton is unstable at some level. In the 1970's, Sheldon Glashow and Howard Georgi proposed the grand unification of the strong, weak, and electromagnetic forces at energies above 1014 GeV. If the ordinary concept of thermal energy applied at such times, it would require a temperature of 1027K for the average particle energy to be 1014 GeV. 4
  5. 5. 5 The unification of the strong force is well beyond our reach at the present time, and the unification of gravity with the other three is out of reach for earthbound experiments. This has led to greater cooperation between high-energy particle physicists and astrophysicists as each group realizes that some of their answers can only come from the other. Ordinary Mechanical Systems: Ordinary mechanical system involves only two forces. (1)Gravity (2) Electromagnetism The gravitational force is apparent property in the Earth’s attraction for objects, which gives them weight. All other forces are electromagnetism in origin. I. Contact forces II. Viscous forces III. Tensile forces IV. Elastic forces I. Contact forces: There are two kinds of contact forces: a. Normal Force b. Frictional Force a. Normal force: Normal force is produced in bodies when one body pushes on another. The normal force is defined as the net force compressing two parallel surfaces together; its direction is perpendicular to the surfaces. b. Frictional force: Frictional force Produced b/w bodies when one surface rubs against another. 5
  6. 6. 6 II Viscous forces: Air resistance. III Tensile forces: Tensile forces are produced in a stretched rope or string. Tension is a reaction force applied by a stretched string (rope or a similar object) on the objects which stretch it. The direction of the force of tension is parallel to the string, away from the object exerting the stretching force. So if an object hangs from a rope due to gravity, the gravitational force on the object points downward, and there is an equal tension force in the rope point upward, making the net force on the object equal to zero. Tension exists also inside the string itself: if the string is considered to be composed of two parts, tension is the force which the two parts of the string apply on each other. The amount of tension in the string determines whether it will break, as well as its vibrational properties, which are used in musical instruments. IV Elastic forces: Elastic forces are that kind of forces arising from the deformation of a solid body which depends only on the body's instantaneous deformation and not on its previous history, and which is conservative. These are produced in a spring 6
  7. 7. 7 6.2 FRICTIONAL FORCES: When one body slides over the surface of another body, an opposing force is set up to resist the motion. This force which opposes the motion is called “force of friction” or friction. Friction is the resistive force acting between bodies that tends to oppose and damp out motion. Friction is usually distinguished as being either 1. static friction (the frictional force opposing placing a body at rest into motion) and 2. Kinetic friction (the frictional force tending to slow a body in motion). 3. In general, static friction is greater than kinetic friction. The friction force is electromagnetic in origin: atoms of one surface "stick" to atoms of the other briefly before snapping apart, causing atomic vibrations, and thus transforming the work needed to maintain the sliding into heat. The study of friction is called tribology. The frictional forces on each body are in a direction opposite to its motion relative to the other body. Frictional forces automatically oppose this relative motion and never aid it even when there is no relative motion, frictional forces may exist b/w surfaces. 7
  8. 8. 8 Factors on which force of friction depends: 1- Normal reaction forces 2- Nature of surface 1- Normal reaction ( R ): Force of friction is directly proportional to normal reaction “ R “ which acts in upward direction against the weight of the body sliding on a surface. The normal force is defined as the net force compressing two parallel surfaces together; its direction is perpendicular to the surfaces. In the simple case of a mass resting on a horizontal surface, the only component of the normal force is the force due to gravity, where FN=mg. In this case, the magnitude of the friction force is the product of the mass of the object, the acceleration due to gravity, and the coefficient of friction. However, the coefficient of friction is not a function of mass or volume; it depends only on the material. For instance, a large aluminum block will have the same coefficient of friction as a small aluminum block. However, the magnitude of the friction force itself will depend on the normal force, and hence the mass of the block. If an object is on a level surface and the force tending to cause it to slide is horizontal, the normal force N between the object and the surface is just its weight, which is equal to its mass multiplied by the acceleration due to earth's gravity, g. If the object is on a tilted surface such as an inclined plane, the normal force is less, because less of the force of gravity is perpendicular to the face of the plane. Therefore, the normal force, and ultimately the frictional force, is determined using vector analysis, usually via a free body diagram. Depending on the situation, the calculation of the normal force may include forces other than gravity. 2-Nature of surface: 8
  9. 9. 9 Force of friction also depends on the nature of two surfaces in contact with each other. And it is constant for various pair of surfaces in contact and its value is known as Co-efficient of friction. Friction is very important in our daily lives. Advantages of frictional forces: 1- In the absence of frictional forces b/w ground & our feet we can not walk. 2- The frictional force b/w the roads & the driving wheels propel an automobile the frictional force prevents the tier from slipping & provides the necessary propelling force. 3- Proper forces of frictional are maintained in the joints of the body to facilitate running or rapid movement of joints in the absence of friction we can not stand upright. Disadvantages of frictional forces: In a car engine oil under pressure is supplied continuously to all bearing surfaces. Absence of oil will allow metal to metal contact & the resulting friction will raise the temperature and cause the bearing and position to seize up. Force Law for Frictional Forces: Now if we want to know the force law for frictional forces or we want to know how to express frictional forces in terms of the properties of the body and its environment. The force law for dry, sliding friction is empirical in character & approximate in their predictions. They do not have the elegant simplicity and accuracy as for the gravitational force law & for the electrostatic force law. Consider a block at rest on a horizontal table ( a ) attach a spring to it to measure the horizontal force ( F ) required to set the block in motion. 9
  10. 10. 10 We find that the block will not move even though we apply a small force fig ( b ). It means that applied force is balanced by an opposing frictional force “ f “ exerted on the block by the table, acting along the surface of contact. As we increase the applied force fig (c) (d) then there is a definite force at which the block will “brake away “from the surface & begins to accelerate. Fig ( e ). By reducing the force once motion has started, it is possible to keep the block in uniform motion without acceleration. Fig ( f ). 10
  11. 11. 11 TYPES OF FRICTION: 1- Static friction 2- Kinetic friction 1- Static friction: The frictional forces existing b/w surfaces at rest with respect to each other are called forces of static friction. A motionless body is subject to static friction. The direction of the static friction force can be visualized as directly opposed to the force that would otherwise cause motion, were it not for the static friction preventing motion. In this case, the friction force exactly cancels the applied force, so the net force given by the Vector sum, equals zero. It is important to note that in all cases, Newton's first law of motion holds. 2-Kinetic friction: The forces acting b/w surfaces in relative motion are called forces of kinetic friction. The friction force is directed in the opposite direction of the Resultant force acting on a body. In the case of kinetic friction, the direction of the friction force may or may not match the direction of motion: a block sliding atop a table with rectilinear motion is subject to friction directed along the line of motion; an automobile making a turn is subject to friction acting perpendicular to the line of motion (in which case it is said to be 'normal' to it). LAWS OF Static FRICITION: These laws were discovered experimentally by Leonardo da vinci. (1452- 1519) Two laws of friction about the maximum force of Static friction b/w any pair of dry un- lubricated surfaces. 1. It is approximately independent of the area of contact over wide limits. 2. It is proportional to the normal force. 11
  12. 12. 12 Normal force / loading force: It is the force which arises from the elastic properties of the bodies in contact. For a block resting on a horizontal table or sliding along it, the normal force is equal in magnitude to the weight of the block. Laws of Kinetic Friction: Two laws of friction about of kinetic friction fk b/w dry un-lubricated surfaces: 1. It is approximately independent of the area of contact over wide limits. 2. It is proportional to the normal force Coefficient of friction The coefficient of friction is a dimensionless quantity used to calculate the force of friction (static or kinetic). While it is often stated that the coefficient of friction (COF) is a "material property," it is better categorized as a "system property." Unlike true material properties (such as conductivity, dielectric constant, yield strength), the COF for any two materials depends on system variables like temperature, speed, atmosphere, as well as on geometric properties of the interface between the materials. For example, a copper pin sliding against a thick copper plate can have a COF that varies from 0.6 at low speeds (metal sliding against metal) to below 0.2 at high speeds when the copper surface begins to melt due to frictional heating. The latter speed, of course, does not determine the COF uniquely; if the pin diameter is increased so that the frictional heating is removed rapidly, the temperature will drop, the pin remains solid and the COF rises to that of a 'low speed' test. 1. Coefficient of Static friction 2. Coefficient of kinetic friction 12
  13. 13. 13 1. Coefficient of Static friction The coefficient of static friction is defined as the ratio of the maximum static friction force (F) between the surfaces in contact to the normal (N) force. Co-efficient of Static friction: ( μs ): The ratio of magnitude of the maximum force of static friction to the magnitude of the normal force is called the co-efficient of static friction for the surfaces involved. f S ≤ µS N −−−−−− →(1) fS µS = N Where f S = magnitude of the force of static friction µS = co-efficient of static friction N = magnitude of normal force NOTE : The equality sign hold only when fS has its maximum value. substance wood on wood 0.25-0.50 steel on steel 0.58 glass on glass 0.9-1.0 Muhammad Yousuf Soomro (Lecturer) Institute of Physics 13
  14. 14. 14 2. Coefficient of kinetic friction The ratio of the magnitude of the force of kinetic friction to the magnitude of the normal force is called the co-efficient of kinetic friction. f K = µ K N − − − − − − → (2) fK µK = N Where f K = magnitude of the force of kinetic friction µK = co-efficient of kinetic friction N = magnitude of the normal force NOTE : The force of the kinetic friction is also reasonably impendent of the relative speed with which the surfaces move over each other. Some values for common substances are given in the following table substance T( ) hickory on dry snow -- 0.08 ice 0 0.020 ice -10 0.035 brass on ice 0 0.020 NOTE : 14
  15. 15. 15 Equations (1) & equation (2) are relations b/w the magnitude only of the normal & frictional forces these forces are always directed perpendicularly to one another IMPORTANT : 1- Both µS , µK ,are dimensionally constant , each being the ratio of ( magnitude) two forces. 2- Usually for a given pair of surfaces µS > µK 3- The actual values of µS & µK depend on the nature of both the surfaces in contact. 4- Both µS , µK can exceed unity, although commonly they are less than 1. NOTE : 1. Both static and kinetic coefficients of friction depend on the pair of surfaces in contact. 2. Their values are usually determined experimentally. 3. For a given pair of surfaces, the coefficient of static friction is larger than that of kinetic friction. A free-body diagram of a block resting on a rough inclined plane, with its weight (W), normal force (N) and friction (F) shown. 15
  16. 16. 16 Sliding Coefficient of Friction when the tangential force F overcomes the frictional force between two surfaces then the surfaces begins to slide relative to each other. In the case of a body resting on a flat surface the body starts to move. The sliding frictional resistance is normally different to the static frictional resistance. The coefficient of sliding friction is expressed using the same formula as the static coefficient and is generally lower than the static coefficient of friction.. Rolling Friction When a cylinder rolls on a surface the force resisting motion is termed rolling friction. Rolling friction is generally considerably less than sliding friction. If W is the weight of the cylinder converted to force, or the force between the cylinder and the flat surface, and R is radius of the cylinder and F is the force required to overcome the rolling friction then. W F =f × R “ f “ is the coefficient of rolling friction and has the same unit of length as the radius R. Typical values for f are listed below Surfaces Rolling Friction f Steel on Steel 0,0005m Wood on Steel 0,0012m Wood on Wood 0,0015m Iron on iron 0,00051m Iron on granite 0,0021m Iron on Wood 0,0056m Polymer of steel 0,002m Hardrubber on Steel 0,0077m Hardrubber on Concrete 0,01 -0,02m Rubber on Concrete 0,015 -0,035m 16
  17. 17. 17 Factors affecting the friction between surfaces Dry surfaces 1. For low surface pressures the friction is directly proportional to the pressure between the surfaces. As the pressure rises the friction factor rises slightly. At very high pressure the friction factor then quickly increases to seizing 2. For low surface pressures the coefficient of friction is independent of surface area. 3. At low velocities the friction is independent of the relative surface velocity. At higher velocities the coefficient of friction decreases. Well lubricated surfaces 1. The friction resistance is almost independent of the specific pressure between the surfaces. 2. At low pressures the friction varies directly as the relative surface speed 3. At high pressures the friction is high at low velocities falling as the velocity increases to a minimum at about 0,6m/s. The friction then rises in proportion the velocity 2. 4. The friction is not so dependent of the surface materials 5. The friction is related to the temperature which affects the viscosity of the lubricant Muhammad Yousuf Soomro (Lecturer) Institute of Physics University of Sindh Jamshoro 17
  18. 18. 18 6-3 The Dynamics Of Uniform Circular Motion If a particle moves at constant speed in a circular path both the velocity and acceleration are constant in magnitude but both change their direction continuously this motion is called uniform circular motion. Examples of uniform circular motion: 1- In atoms, the electrons move in circular orbits around the nucleus. 2- The earth and others plants keep as moving around the sun in fixed orbits. 3- Computer disks. If a body is moving at uniform speed “ v ” in a circle or a circular arc of radius “r” it v2 experience a centripetal acceleration “a” whose magnitude is . r The direction of centripetal acceleration is always radially inward towards the center of circle. This centripetal acceleration is a vector because even though its magnitude remains constant, its direction changes continuously as the motion progresses. If a body undergoing uniform circular motion, then a net force must be acting on it & the magnitude of net force mv 2 ∑ F = ma = r The direction of net force acting at any instant must be the direction of centripetal acceleration “a“ at that point, radially inward. 18
  19. 19. 19 Example : A disk is moving with uniform circular motion & it moving on the end of a string in a circle on a frictionless horizontal table as shown in figure. The net force on the disk is provided by the tension “T“in the string. It string accelerates the disk by constantly changing the direction of its velocity so that the disk moves in s circle. The direction of T is always towards the pin at the center, and its magnitude must equal mv 2 to r mv 2 T= R If the string were to be cut where it joins the disk, there would be no net force exerted on the disk. The disk would then move with constant speed in a straight line along the direction of the tangent to the circle at the point at which the string was cut. 19
  20. 20. 20 Centripetal Forces: Forces responsible for uniform circular motion are called centripetal forces because they are directed towards the center of the circular motion. So 1- for the revolving disk , the centripetal force is a tensile provided by the string. 2- For the Moon revolving around the earth the centripetal force is the gravitational pull of the earth on the Moon. 3- For an electron circulating about an atomic nucleus the centripetal force is electrostatic. Examples of Centripetal forces: The Conical Pendulum: Consider a small body of mass “ m “ revolving in a horizontal circle with constant speed “ v” at the end of a string of length “ L “. As the body swings around, the string sweeps over the surface of an imaginary cone. This device is called a Conical Pendulum. We want to find the time required for one complete revolution of the body. If the string makes an angle θ with the vertical, the radius of the circular path is R = L sin θ 20
  21. 21. 21 The forces acting on the body of mass “ m “ are its ( i ) Weight mg ( ii ) Tension T Of the string. Apply Newton’s second law of motion ∑F = T + mg ∑ F = mg ∴g = a Or ∑F = ma We can resolve T at any instant into a radial and vertical components Tr = −T sin θ (Radial component) Tr = T cos θ (Vertical component) The radial component is –ve because if radial direction to be positive outward from axis. 21
  22. 22. 22 As there is no vertical acceleration in body so Z- component of Newton’s second law of motion ∑ F = T − mg = 0 z z Or T cos θ =mg −−−−−−→(1) The radial acceleration is V2 ar =− R It is –ve because it acts radially inward. Radial acceleration is supplied by Tr which provides the centripetal force acting on “m “. From the radial component of Newton’s second law of motion ∑F r = Tr = mar V2 ∑ F = −T sin θ = m(− R ) mV 2 −T sin θ = − R mV 2 T sin θ = − − − − − − → (2) R Muhammad Yousuf Soomro (Lecturer) Institute of Physics University of Sindh Jamshoro 22
  23. 23. 23 Dividing equation ( 2 ) by equation ( 1 ) T sin θ mV 2 R = T cos θ mg sin θ =V 2 R cos θ g V2 tan θ = Rg V 2 =tan θRg V = Rg tan θ This equation gives the constant aped of the body. Period of Motion: Now if t represents the time for one complete revolution of the body . 23
  24. 24. 24 2π R V = t or 2π R t = V 2π R t = Rg tan θ Squaring both sides 4π R 2 2 t 2 = Rg tan θ 4π R 2 t 2 = g tan θ 4π R2 t 2 = g tan θ R t = π 2 g tan θ As R =l sin θ 24
  25. 25. 25 L sin θ t =π 2 g tan θ L sin θ t =π sin θ 2 g( ) cos θ L cos θ t =π 2 g Note : This equation gives the relation b/w t, L, θ. T is called period of motion and it does not depend on mass m. 6-4 Equations & Motion Constant and non constant forces Constant forces: Forces that dose not depend on time, velocity or position. Examples : 1. Object falling near the earth surface. 2. Braking car If force is constant, then the acceleration is constant, and for constant acceleration the solution in one dimensional for V(t) & x(t) are easily obtained. Constant forces demonstrate the applications of Newton’s law and they are certainly easier to work with than non constant forces. 25
  26. 26. 26 Many particle problems often include forces that remain constant under many circumstances. Examples : 1 Gravity near the earth surface 2 Frictional forces 3 Tension forces in a strings Find V(t) for constant force: Let “ a “ represent constant acceleration & an object starts with velocity “ Vo “ at time t = 0 & at time “ t “ later it has velocity V. V2 − V1 ∆V a= = t2 − t1 ∆t Or 26
  27. 27. 27 ∆V V − V0 a= = ∆t t −0 V − V0 a= t at = V − V0 V = V0 + at − − − − − − → (1) This important result allows us to find the velocity at all later times. This equation No. 01 shows the velocity as a function of time, and it can be written as V(t). Note : Equation 01 is in the form of Y= mx+b, which describes the graph a straight line. Non-Constant forces: Forces that depends on time, velocity & position. Examples : 1 Swinging pendulum bob 2 Rocket blasting toward earth’s orbit 3 A rain drop falling against air resistance. 27
  28. 28. 28 When the force is not constant then we can not use constant acceleration formulae to find V(t) & x(t). Methods of solving problems: 1 Analytical method 2 Numerical method 1 Analytical method: This method involving integral calculus. Find V( t ): As V a = t or dv a = dt a.dt =dv dv = .dt a Take integration both sides with limits of velocity V as follows Vo at time t = 0 & V at time t = t Limits of time from 0 to t 28
  29. 29. 29 V 1 ∫ dv =∫a.dt Vo 0 V 1 ∫ dv =a ∫dt Vo 0 [V ]Vo =a [ t ] 0 V 1 Appling limits V − Vo = a (t − 0) V − Vo = at V = Vo + at or V (t ) = Vo + at 29
  30. 30. 30 Find x ( t ): As S V = S =x t or x V = t dx V = dt dx = V .dt dx = (Vo + at )dt dx = Vo.dt + a.t.dt Integrate with limits from position xo at time 0 to position x at time t x t t ∫ dx = ∫V xo 0 0 dt + ∫ at dt 0 x t t ∫ dx = V ∫ dt + a ∫ tdt x0 0 0 0 t t 2  [ x] x = V0 [ t ] 0 + a   x t 0  2 0 30
  31. 31. 31 Apply limits ( x − x0 ) = V0 (t − 0) + a( 1 t 2 ) 2 x − x0 = V0 t + 1 at 2 2 x = x0 + V0 + 1 at 2 2 or x(t ) = x0 + V0 t + 1 at 2 2 Physical phenomenon involving non-constant forces: 1. Forces depending on the time 2. Forces depending on the velocity 3. Forces depending on the position 2-Numerical methods: In this method we use a computer to evaluate the integrals, obtaining not the analytic functions V( t ) & x( t ) but instead the numerical values of V & x at any time t this can be done to any desired level of precision. Muhammad Yousuf Soomro (Lecturer) Institute of Physics University of Sindh Jamshoro 31
  32. 32. 32 Time-Dependent Forces: (analytical methods) When the forces is time dependent, we can obtain an acceleration a(t) using Newton’s laws of motion. Find V( t ): Find the velocity by direct integration. V as a= t dv a= dt dv = a (t ) dt Integrate with limits from Time t = 0 , when the initial velocity is Vo , to time t when velocity is V V t ∫ dv= ∫a(t)dt V0 0 t [ V]V = ∫a(t)dt V 0 0 t V-V0 = ∫a(t)dt 0 t V=V0 + ∫a(t)dt 0 or t V(t)=V0 + ∫a(t)dt 0 32
  33. 33. 33 Find x ( t ): as V = dx dt dx = V dt or dx = V (t ) dt Integrate with limits time t = 0 when the particle is located at xo, to time t = t, when the position is x . x t ∫ dx= ∫V (t)dt x0 0 t [ x] x = ∫V (t)dt x 0 0 t x-x 0 = ∫V (t)dt 0 t x=x 0 + ∫V (t)dt 0 or t x (t)=x 0 + ∫V (t)dt 0 33
  34. 34. 34 x t x t ∫ dx = ∫ V (t )dt x0 0 ∫ dx =∫V (t )dt x0 0 t t [ x] x = ∫ V (t )dt x [ x ] x =∫V (t ) dt x 0 0 0 0 t t x − x0 = ∫ V (t ) dt x −x0 = ∫V (t ) dt 0 0 t t x = x0 + ∫ V (t ) dt x = x0 +∫V (t ) dt 0 0 or or t t x (t ) = x0 + ∫ V (t ) dt x (t ) = x0 +∫V (t ) dt 0 0 Chapter No. 07 Mohd Yousuf Soomro WORK & ENERGY Institute of Physics Work: When force acts upon a body and if it causes displacement also, then work is said to be done by the force upon the body. Or Dot product of two vector quantities (force & displacement) is called work. Work is a scalar quantity. rr W = F .d Where F = force d = displacement moved by the body in the direction of force. NOTE : Sometimes the force and displacement may not be in the same direction. Examples: 34
  35. 35. 35 A wooden block is pulled by means of a rope, it moves in the horizontal direction, but the force acts along the rope. If the force makes an angle “ θ “ with the direction of motion then work done can be found by the production of the component of the force along the direction of motion and the displacement “ S “ moved by the block. We can express work done by the force analytically as W = F . d cosθ Where r F = magnitude of the vector F r d = magnitude of the vector d r r θ = angle b/w F and d Work is an algebraic quantity. It can be positive, negative , zero depending upon the value of angle b/w force ( F ) and displacement ( d ). Cases of work done: ( i ) case: When the components of force is in the same direction of the displacement ( θ = 0 ) then the work is positive. 35
  36. 36. 36 r r W = F . d cos θ if θ =0 W = F .d cos 0 W = F .d (1) cos00 = 1 W = F .d Example : When a spring is stretched, then work done by the stretched ( stretching force ) is positive. ( ii ) case: When the direction of the force is opposite to the direction of the displacement ( θ =1800 ) the work done is negative. 36
  37. 37. 37 rr W = F .d cos θ if θ = 1800 W = F .d cos1800 cos1800 = − 1 W = F .d (− 1) W = − F .d Example : Work done by the “gravitational force “on the body being lifted is negative. When a body is lifted against gravity at very slow speed , the angle b/w gravity and displacement is 1800. ( iii ) case: When the force acts at right angle to the displacement ( θ = 900 ) THEN THE WORK done is zero. It means that force does not produce work. 37
  38. 38. 38 rr W = F .d cos θ if θ =0 W = F .d cos 900 W = F .d X 0 cos 900 = 0 W =0 Example : work done by the centripetal force is zero because the centripetal force is always at right angles to the direction in which the body moving. Units: SI System: (MKS System) In the S.I system the unit of work is Newton-meter which is commonly known as Joule(J ). 1 Newton. meter = 1 Joule 38
  39. 39. 39 1N. m = 1 J The work of 1 joule is said to be done if a force of 1 Newton acts on a body and if it moves through a distance of 1 meter in the direction of force. CGS system: ( centimeter , gram , second system ) In CGS system the unit of work is erg. 1 erg = 1 dyne 1 cm 1 erg = 10-7 J The work of 1 erg is said to be done if a force of 1 dyne acts on a body and if it moves through a distance of 1 centimeter in the direction of force. FPS System ( foot, pound , second system ) / British Engineering System : In FPS System the unit of work is Foot . Pound ( ft-lb ). 1 ft. lb = 1.355 joule Relationships b/w units: 1 Joule = 105 dynes X 102 cm 1 Joule = 107 ( dynes .cm ) 1 Joule =107 Joule 1055 Joule = 1 Btu (Btu= British thermal unit) NOTE : In the physics of molecules , atoms , and elementary particles a much smaller unit is used , it is called electron volt ( e.V ) 39
  40. 40. 40 1eV = 1.60 X 10−19 J Multiples of the electron volts: 1 million eV = 1 M eV = 106 eV 1 billion eV = 1 B eV = 1012 eV Muhammad Yousuf Soomro Lecturer Institute of Physics University of Physics Jamshoro Physical dimension of Work: As Work = force. displacement W = F . d − − − − − −− → (1) Since F = ma Or F = mass × acceleration L F = M × ( ) M = mass T2 a = acceleration L = length T = time So equation ( 1 ) becomes 40
  41. 41. 41 L W = M× ( 2 )( L) T ML2 W = 2 T W = M L2 T −2 7.1 Constant Force on a Particle: Suppose a particle of mass “m “is subjected to a constant force “ F “. We assume the motion of a particle along x- axis. 41
  42. 42. 42 W = F .d or W = F .dx − − − −− → (∗) As we know from Newton’s second law of motion F = ma dv F =m dt So equation ( * ) becomes dv F . dx = m .dx − − − − − (1) dt Suppose the particle is displaces from “Xo” to “X”. Let the velocity at “ Xo” be “ Vo” and velocity at “ X “ be “ V “. Applying these conditions on equation (1) and integrate it. x v dv ∫ F .dx = ∫ m x0 V0 dt .dx Since the force “ F “ and mass “ m “ are constants so we may take these constants out of the integral sign. x v dv F ∫ dx = m ∫ .dx x0 V0 dt x v dx F ∫ dx = m ∫ .dv x0 V0 dt 42
  43. 43. 43 x v dx F ∫ dx = m ∫ Vdv V= x0 V0 dt 2 v v  F [ x] x = m   x 0  2  v0 Appling the limits 2 v 2 v0 F ( x − x0 ) = m( − ) 2 2 F ( x − x0 ) = 1 mv 2 − 1 mv0 − − − − → (2) 2 2 2 2 We call 1 mv as the Kinetic energy of the body. The R.H.S of equation ( 2 ) is the 2 change in the K.E of the particle when it changes its position from “xo” to “x”. Thus “The work done by a force is equal to the change in K.E of the particle. “ 7.2 Work Done by a Variable Force: The work done by a variable force can not be calculated by the direct use of formula W= F.d for the entire displacement “ d” because the force is continuously changing with the displacement. So to calculate the work done by a variable force , we make two assumptions. (1) The direction of force is along positive X-axis and its magnitude is changing with displacement. (2) The body is constrained to move in positive X-axis direction only. Suppose a body of mass “ m “ is moving under a variable force “ f “. The value of force “ f” is different at different positions. 43
  44. 44. 44 We assume that the body moves from its initial position “ A “ to the final position “ B “ in the direction of the variable force. If the body has moved from position “ A “ to the position “ B “ then we divide the total displacement “ AB “ into small interval each of width “Δr “ . During any short displacement interval “Δr “the force remains almost constant. Thus the force for any one displacement interval = dot product of “F “and “Δr “. The total work done for the displacement “ AB “ WA→ B = F1∆r1 + F2 ∆r2 + F3∆r3 + ........ or B W A→ B = ∑ F ( r ) ∆ r A The constant force is taken to be the force which acts at the beginning of each interval. Let the total number of intervals from “A” to “ B “ be “ N “. Now if the number of intervals are made infinite ( N → ∞ ) and the width “ Δr “ is made so small that it tend to zero ( ∆r → 0 ) 44
  45. 45. 45 B WA→ B = ∆r → 0∑ F .∆r A Now B B ∆r → 0 ∑ F .∆r = ∫ F .dr A A therefore the total force B WA→B = ∫ F .dr − − − −− → (1) A As we known that F=ma dv dv F=m a= dt dt So equation ( 1 ) becomes B dv WA→B = ∫ m dr − − − −− → (2) A dt Since “ m “ mass is constant so we may take it out of the ingral sign B dv WA→B = m ∫ dr A dt B dr WA→B = m ∫ dv A dt B WA→B = m ∫ Vdv − − − − → (3) A Integrate equation ( 3 ) with respect to ”v” 45
  46. 46. 46 B  2 V =m    2 A B 1  = mV 2  2 A Applying limits 1 1 mVB2 − mVA 2 2 2 It means K .E at B =K .E at A 1 1 mVB2 = mVA −− −− → (4) 2 2 2 Example : An example of a variable force involving in the motion of spring. We consider a spring that acts on a particle of mass “m “ . The particle moves in the horizontal direction with the origin ( x = 0 ). 46
  47. 47. 47 Let the particle be displaced a distance “ x “ from its original position x = 0 , as the agent exerts a force “ F “ on the particle , the spring exerts an opposing force “ F “. We know from the Hook’s law that “Force acting on the body is directly proportional to the displacement of the body”. F = − kx Where K= force constant or spring constant. Let us consider the work done on the particle by the spring when the particle moves from initial position “ xi “ to “ xf”. F = −k x W = F .d W = ( −kx ).dx xf W = ∫ ( −kx ).dx xi As k is constant so we may take it out of the integral sign. xf W = k ∫ (− x).dx − − − − → (1) xi Integrating equation ( 1 ) with respect to “ x”. 47
  48. 48. 48 xf  x  2 W = k −   2 xi Now applying limits kx 2 kxi2 f W =− + 2 2 2 kxi2 kx f W= − − − − − → (2) 2 2 7-4 Work Energy & The Work Energy Theorem: Work energy theorem: The net work done by the forces acting on a particle is equal to the change in the Kinetic energy of the particle. Wnet = K f − Ki = ∆ K Explanation : Consider a net work done (W net) on a particle not by single force but all the forces that act on the particle. Ways to find Net Work: There are two ways to find the net work. 1- Find the net force: It means the vector sum of all the forces that acts on the particle. Fnet = F1 + F2 + F3 + ....... 48
  49. 49. 49 Then treat this force as a single force in calculating the net work by following equation. xf W = ∫ F ( x)dx (One dimension) xi f f W = ∫ F .ds = ∫ F cos φ ds (Many dimensions) i i Muhammad Yousuf Soomro Lecturer Institute of Physics University of Sindh Jamshoro 2- Calculate the work done by each of the forces that acts on the particle: W1 = ∫ F1.ds W2 = ∫ F2 .ds W3 = ∫ F3 .ds Wn = ∫ Fn .ds To find net work, add the work done by each of the individual forces. Wnet = W1 + W2 + W3 + ......... We know that a net unbalanced force applied to a particle will change its state of motion by accelerating it from initial velocity “ Vi “ to final velocity “ Vf “. Under the influence of constant force , the particle moves from Xi to Xf & it accelerates uniformly from Vi to Vf. The work done is 49
  50. 50. 50 Wnet =Fnet ( x f −xi ) Wnet =ma ( x f −xi ) Because acceleration is constant we use the equation V 2 = V02 + 2a ( x − x0 ) or V f2 = Vi 2 + 2a ( x f − xi ) 1 1 Wnet = mv 2 − mv f 2 2 It means: “The result of the net work on the particle has been to bring about a change in the value of quantity 1 mv 2 from point “ i “ to point “ f “. This quantity is called the Kinetic 2 energy K of the particle. K = 1 mv 2 2 NOTE : Work energy theorem holds for both constant forces as well as non constant forces: The work energy theorem does not represent a new independent law of classical mechanics. It is useful for solving problems in which the net work done on a particle by external forces is easily computed and in which we are interested in finding the particle’s speed at certain positions. General Proof of Work-Energy Theorem: 50
  51. 51. 51 Let Fnet represent the net forces acting on the particle. The net work done by all the external forces that act on the particle is Wnet = ∫ Fnet dx − − − −− ⇒ (1) Fnet = ma As dv dv Fnet = m ∴a= dt dt By using chain rule dv dx Fnet = m dx dt dv Fnet = m v dx dv Fnet = mv − − − −− ⇒ (2) dx Put equation ( 2 ) in equation ( 1 ) 51
  52. 52. 52 Wnet = ∫ Fnet dx or dv Wnet = ∫ mv dx dx Wnet = ∫ mv dv Integrate from initial velocity “Vi” to final Velocity “Vf” vf Wnet = ∫ mvdv vi vf Wnet = m ∫ vdv vi 2 vf v  Wnet = m    2  vi 52
  53. 53. 53 Applying limits Wnet = m( 1 v 2 − 1 vi2 ) 2 f 2 Wnet = 1 m(v 2 − vi2 ) 2 f or Wnet = 1 mv 2 − 1 mvi2 2 f 2 This shows that the work – energy theorem holds also for non constant forces. 7-5 Power “The work done in unit time called power”. Or “The rate of doing work is called power “. F .d Power = t W Power = ∴ W = F .d t d Power = F .V ∴ =V t Kinds of Power: 53
  54. 54. 54 1. Average power 2. Instantaneous power 1. Average Power: If “Δw“ is the work done in a time “ Δt “ the power is called average power and it is written as Pav. ∆W P = av ∆ t 2. Instantaneous power: ∆W If ∆t → 0 the limiting value of is called Instantaneous power at time “ t “. ∆t ∆W P = lim − − − −− → (1) ∆t →0 ∆t As we know that W = F .d So equation ( 1 ) becomes dr P = lim F . ∆t →0 dt P = F .V It means power is equal to the dot product of force and velocity. Units: SI System: 54
  55. 55. 55 The S.U unit of power is Joule per Second (J/s) which is also called Watt (W). This unit is named in honor of James Watt (1736-1819) 1 W = 1 J / s = 1 Kg.m2 / s3 Watt: One watt is defined as: “The rate of doing work or using energy at Joule per Second”. Bigger Units: The bigger units of power are Kilowatt ( kW) Mega watt ( MW) and Giga Watt ( Gw ) 1 kW= 103 watt = 1000 w 1 Mw=106 watt=1000000w=1000 Kw 1 Gw= 109 watt=1000000000w FPS System: In British Engineering system the unit of power is ft-lb / sec (foot. Pound / second ). Since this unit is quite small, therefore a bigger unit called Horse Power (hp) is used. 1 horsepower = 745.7 watts The term "horsepower" was coined by the engineer James Watt (1736 to 1819) in 1782 while working on improving the performance of steam engines. Horse Power: “One horse power (hp) is the power of an agency which does work at the rate of 550 ft-lb/sec or 33000 ft-lb per minute. 550 ft-lb/sec = 33000 ft-lb / min 1 horsepower = 745.7 watts Or 1 hp = 746 watt 55
  56. 56. 56 1 min = 60 seconds 1 ft = 0.3048 m and 1 lb = 0.45359237 kg Relationship b/w Kilowatt hour & Joule: The term kilowatt hour (KWh ) is originated from the unit of work. One Kilo watt is defined as the work done in one hour by an agency working at the constant rate of 1 Kw that is 1000 Joule per Second. 1 hour = 3600 seconds 1 Kwh = 1000 X 3600 = 3.6 X 106 Joules Relationship b/w Horse power & Watt: 1 hp= 550 ft –lb / sec We know that 1 ft = 0.3048 m 1 lb = 4.448 N So 1 hp = 550 X 1.351 X 4.448 = 550 X 1.351 N.m / sec = 550 X 1.351 Joules / sec 1 hp = 746 watt Some Important Units of Energy MECHANICAL ENERGY: 56
  57. 57. 57 Metric Units: SI: Joule (J) 1 J = 1 N-m English Units: foot-pound (ft-lbs) 1 ft-lbs = 1.356 J HEAT ENERGY: Calorie (Cal) 1 Cal = 4.186 J British Thermal Unit (BTU) 1 BTU = 1,055 J ELECTRICAL & ATOMIC ENERGY: Electron Volt (eV) 1 eV = 1.6x10-19 J Kilowatt-hour (kWh) 1 kWh = 3.6x10+6 J Muhammad Yousuf Soomro Lecturer Institute of Physics University of Sindh Jamshoro 57