Lect.3 Properties of Matter Ch2+Ch3 PDF

Properties of Matter Dr/ Reham Aly Lecturer Faculty of Science - Helwan University 1  Contents Chapter 1: Physical Quantities and Measurement. Chapter 2: Work and Kinetic Energy. Chapter 3...

Properties of Matter Dr/ Reham Aly Lecturer Faculty of Science - Helwan University 1  Contents Chapter 1: Physical Quantities and Measurement. Chapter 2: Work and Kinetic Energy. Chapter 3: Mechanical Properties of Materials. Chapter 4: Fluid Mechanics. Chapter 5: The Law of Universal Gravitation. Chapter 6: Sound Waves in Elastic Media. 2  Lecture 3: Work and Kinetic Energy ØWork Done by a Constant Force. ØWork Done by a Varying Force. ØWork Done by a Spring. Prev. Lec ØKinetic Energy and the Work – Kinetic Energy Theorem. ØSituations Involving Kinetic Friction. ØPower. ØConservation and Non-conservation Forces. Today ØConservation of Mechanical Energy. ØRelationship Between Conservative Forces and Potential Energy. 3  Situations Involving Kinetic Friction Suppose a book moving on a horizontal surface is given an initial horizontal velocity 𝑣! and slides a distance 𝑑 before reaching a final velocity 𝑣". The external force that causes the book to undergo an acceleration in the negative 𝑥 direction is the force of kinetic friction 𝑓# acting to the left, opposite the motion. $ $ The initial kinetic energy of the book is % 𝑚𝑣!%, and its final kinetic energy is % 𝑚𝑣"%. By applying Newton’s second law, the only force acting on the book in the x-direction is the friction force. & 𝐹 = 𝑚𝑎 𝐹& − 𝑓# = 𝑚𝑎& 4  Situations Involving Kinetic Friction By using Kinematics equation for motion under constant acceleration: # # 𝑣!" − 𝑣!$ = 2𝑎! 𝑑 1 # # %&'($)'*$+, -* % 1 # # 𝑎! 𝑑 = 𝑣!" − 𝑣!$ 𝑚𝑎! 𝑑 = 𝑚 𝑣!" − 𝑣!$ 2 2 %&'($)'*$+, -* / 𝐹! − 𝑓. = 𝑚𝑎! 𝐹! 𝑑 − 𝑓. 𝑑 = 𝑚𝑎! 𝑑 1 # # 𝐹! 𝑑 − 𝑓. 𝑑 = 𝑚 𝑣!" − 𝑣!$ 2 ∴ 𝐹! 𝑑 − 𝑓. 𝑑 = ∆𝐾 5  Situations Involving Kinetic Friction Part of this lost Kinetic energy goes into warming up the book, and the rest goes into warming up the surface over which the book slides. When friction as well as other forces acts on the object, the Work-Kinetic theorem: 𝐾! + & 𝑊'()*+ − 𝑓# 𝑑 = 𝐾" where ∑ 𝑊'()*+ represents the sum of the amounts of work done on the object by forces other than kinetic friction. 6 q Friction force 𝑓. The force that resists motion when the surface of one object in contact with the surface of another. 𝑓. = 𝜇 𝑛 where: 𝜇: coefficient of friction. 𝑛: normal force. 𝑛 = 𝑚𝑔 7  Situations Involving Kinetic Friction qExercise: A Block Pulled on a Rough Surface A 6.0-kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of 12 N. Find the final speed of the block after it has moved 3.0 m, if the surface is not frictionless but instead has a coefficient of kinetic friction of 0.15. Solution Normal force (n) balance force of gravity and these vertical forces does no work on the block because the displacement is horizontal. There is friction force 𝑓! = 𝜇 𝑛 = 0.15 6.0 𝑘𝑔 9.8 𝑚/𝑠 ". −𝑓! 𝑑 = − 8.82 𝑁 3.0 𝑚 = −26.6 𝐽 𝑓# The final speed of the block: 1 1 𝑚𝑣# + D 𝑊$%&'( − 𝑓! 𝑑 = 𝑚𝑣)" " 2 2 1 0 + 36 𝐽 − 26.6 𝐽 = 6.0 𝑘𝑔 𝑣)" 𝑣) = 1.8 𝑚/𝑠 2 8  Situations Involving Kinetic Friction qExercise: A Block Pulled on a Rough Surface A 6.0-kg block initially at rest is pulled to the right along a horizontal not frictionless surface by a constant horizontal force of 12 N. Find the acceleration of the block from Newton’s second law and 𝟐 determine its final speed of the block after it has moved 3.0 m. Using the kinematics equation 𝐯𝒙𝒇 𝟐 − 𝐯𝒙𝒊 = 𝟐𝒂𝒙 𝒅. surface is not frictionless but instead has a coefficient of kinetic friction of 0.15. Solution By applying newton’s second law: ' 𝐹% = 𝑚𝑎% 𝑓# 𝐹 − 𝑓& = 𝑚𝑎% 12 − 0.15 6.0𝑘𝑔 9.8 𝑚⁄𝑠 ' = 6.0𝑘𝑔 𝑎% 𝑎% = 0.53 𝑚/𝑠 ' ' ' Using kinematics equation: 𝑣%( − 𝑣%) = 2𝑎% 𝑑 ' 𝑚 𝑣%( − 0 = 2 0.53 ' 3.0 𝑚 𝑣%( = 1.8 𝑚/𝑠 𝑠 9  Power If the work done by the force in the time interval ∆t is 𝑊, then the average power expended during this interval is defined as the total work done over a given time interval. ∆𝑊 𝑃= ∆𝑡 Power is the time rate of energy transfer, the most general expression. ∆𝐸 𝑃= ∆𝑡 Instantaneous power as the limiting value of the average power as ∆𝑡 approaches zero: ∆𝑊 𝑑𝑊 𝑃 = lim = ∆(→H ∆𝑡 𝑑𝑡 IJ IK 𝑑𝑊 = 𝐹. 𝑑𝑠 𝑃= I( = 𝐹. I( /0 𝑃 = 𝐹. 𝑣 𝑣= Instantaneous Speed 10 /(  Power q SI unit of Power: Joules/sec = Watt (W) = kg.m2/s3 q In the British engineering system, the unit of Power is horsepower (hp) 1hp = 746 W q Unit of Energy (Work) in terms of Power: Ø 1kWh is the energy converted or consumed in 1 h at the constant rate of 1kW = 1000 J/s Unit of Energy not Power 1kWh = (103 W) (3600 S.)=3.6X106 J When you pay your electric bill, you pay the power company for the total electrical energy you used during the billing period. This energy is the power used multiplied by the time during which it was used. For example, a 300-W light bulb run for 12 h would convert (0.300 kW) (12 h) = 3.6 kWh of electrical energy. 11  Conservative Forces qConservative force: 1. The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle but depends only on the initial and final positions. 2. In other words, the work done by a conservative force is the same for any path connecting two points. 12  Conservative Forces qExample: If we throw a ball in two different paths but with the same start and end points. Ø Let us first throw a ball and find the work done by gravitational force when the ball reaches the height h. ) 𝑊 = 8 −𝑚𝑔 𝑑𝑥 = −𝑚𝑔ℎ H ØLet us say second thrown the same ball with a higher velocity such that the ball passes the height h, travels further, reaches a height of L, then comes back to h, and finally reaches the ground. L ) 𝑊 = 8 −𝑚𝑔 𝑑𝑥 + 8 −𝑚𝑔 𝑑𝑥 = −𝑚𝑔ℎ H L vThe gravitational force is a “conservative force” because it is path-independent. 13  Non-Conservative Forces qNon-Conservative force: 1. The work done by a non-conservative force on a particle moving between any two points is dependent of the path taken by the particle. 2. In other words, the work done by a non-conservative force is not the same for any path connecting two points. 14  Non-Conservative Forces qExample: Suppose you displace a book between two points on a table. If the book is displaced in a straight line along the blue path between points A and B, you do a certain amount of work against the kinetic friction force to keep the book moving at a constant speed. Now, imagine that you push the book along the brown semi-circular path. You perform more work against friction along this longer path than along the straight path. The work done depends on the path. So the friction force non-conservative force. 15  Conservation of Mechanical Energy vThe total mechanical energy of a system remains constant in any isolated system of objects that interact only through conservative forces. The total mechanical energy E of a system is defined as: the sum of the kinetic K and potential energy U. 𝐸 =𝐾+𝑈 According to the principle of conservation of energy 𝐸$ = 𝐸" 𝐾$ + 𝑈$ = 𝐾" + 𝑈" 𝐾" − 𝐾$ + 𝑈" − 𝑈$ = 0 ∆𝐾 + ∆𝑈 = 0 16  Change in Mechanical Energy for Non- conservative Forces If a non-conservative force acts within a system: ∆𝐾 + ∆𝑈 = −𝑓. 𝑑 Mechanical Energy is not conserved due to friction 17  Relationship between Conservative Forces and Potential Energy If a conservative force acts on an object, the object moves along the x axis from 𝒙𝒊 to 𝒙𝒇 , the change in the potential energy of the object equals the negative of the work done by that force: 𝑑𝑈 = −𝐹& 𝑑𝑥 The conservative force is related to the potential energy through the relation: 𝑑𝑈 𝐹& = − 𝑑𝑥 vA conservative force acting on an object equals the negative derivative of the potential energy of the system with respect to x. 18  Relationship between Conservative Forces and Potential Energy qWe can easily check this equation for two examples already discussed. 1 # Potential Energy 𝑈0 = 𝑘𝑥%1! 2 From the relation between Conservative force and potential energy: 𝑑𝑈K 𝑑 1 % 𝐹K = − 𝐹K = − 𝑘𝑥OP& = −𝑘𝑥 𝑑𝑥 𝑑𝑥 2 Restoring Force of Spring  Relationship between Conservative Forces and Potential Energy II. The gravitational potential energy is: 𝑈, = 𝑚𝑔ℎ From the relation between Conservative force and potential energy: 𝑑𝑈, 𝐹, = − 𝑑ℎ 𝑑 𝐹, = − 𝑚𝑔ℎ = −𝑚𝑔 𝑑ℎ 20 Example2.11 Motion on a Curved Track A child of mass m rides on an irregularly curved slide of height h = 2 m as shown in. The child starts from rest at the top. (a) Determine his speed at the bottom, assuming no friction is present. Solution The normal force does no work on the child because it is normal on the child displacement. The only force acting on the child is gravitational force (conservative force). By applying the principle of conservation of mechanical energy: 𝐾! + 𝑈! = 𝐾" + 𝑈" 1 𝑜 + 𝑚𝑔ℎ = 𝑚𝑣"% + 0 2 𝑣" = 2𝑔ℎ = 2 9.8𝑚/𝑠 % 2𝑚 = 6.26𝑚/𝑠 21 (b) If a force of kinetic friction acts on the child, how much mechanical energy does the system lose? Assume that vQ = 3m/s and m=20 kg Solution ∆E = 𝐸" − 𝐸! = 𝐾" + 𝑈" − 𝐾! + 𝑈! 1 ∆E = 𝑚𝑣"% + 0 − 0 + 𝑚𝑔ℎ 2 1 3𝑚 % 9.8𝑚 ∆E = × 20𝑘𝑔 − 20𝑘𝑔 % 2𝑚 = −302𝐽 2 𝑠 𝑠 ∆E is negative because the force of kinetic friction is reducing the mechanical energy of the system 22 Example Ball in Free Fall A ball of mass m is dropped from a height h above the ground, as shown in. (a) Neglecting air resistance, determine the speed of the ball when it is at a height y above the ground. Solution The only force acting on the ball is gravitational force (conservative force). By applying the principle of conservation of mechanical energy: 𝐾! + 𝑈! = 𝐾" + 𝑈" 1 𝑜 + 𝑚𝑔ℎ = 𝑚𝑣"% + 𝑚𝑔𝑦 2 𝑣"% = 2𝑔 ℎ − 𝑦 𝑣" = ± 2𝑔 ℎ − 𝑦 The speed is always positive (scaler). 𝑣" = 2𝑔 ℎ − 𝑦 23  Contents Chapter 1: Physical Quantities and Measurement. Chapter 2: Work and Kinetic Energy. Chapter 3: Mechanical Properties of Materials. Chapter 4: Fluid Mechanics. Chapter 5: The Law of Universal Gravitation. Chapter 6: Sound Waves in Elastic Media. 24  Ch. 3: Mechanical Properties of Materials qContents: 1. Introduction. 2. Young’s Modulus: Elasticity in Length. 3. Shear Modulus: Elasticity of shape. 4. Bulk Modulus: Volume Elasticity. 5. Prestressed concrete. 25  Introduction All objects are deformable to some extent. It is possible to change the shape or the size (or both) of an object by applying external forces. As these changes take place, internal forces in the object resist the deformation. We shall discuss the deformation of solids in terms of the concepts of stress and strain. stress is the external force acting on an object per unit cross sectional area. The result of a stress is strain, which is a measure of the degree of deformation. For sufficiently small stresses, stress is proportional to strain; the constant of proportionality depends on the material being deformed and on the nature of the deformation. 26  Introduction We call this proportionality constant the elastic modulus. The elastic modulus is defined as the ratio of the stress to the resulting strain: 𝑠𝑡𝑟𝑒𝑠𝑠 𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 = 𝑠𝑡𝑟𝑎𝑖𝑛 We consider three types of deformation and define an Elastic Modulus for each: 1. Young’s modulus measures the resistance of a solid to a change in its length. 2. Shear modulus measures the resistance to motion of the planes within a solid parallel to each other. 3. Bulk modulus measures the resistance of solids or liquids to changes in their volume. 27  Young’s Modulus: Elasticity in Length Long bar of cross-sectional area 𝐴 and initial length 𝐿! that is clamped at one end. When an external force is applied perpendicular to the cross section, internal molecular forces in the bar resist distortion (“stretching”). But the bar reaches an equilibrium situation in which its final length 𝐿" is greater than 𝐿!. In which the external force is exactly balanced by the internal forces. In such a situation, the bar is said to be stressed. 28  Young’s Modulus: Elasticity in Length We define the tensile stress as the ratio of the magnitude of the external force F to the cross- sectional area A, where the cross section is perpendicular to the force vector. 𝐹 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝐴 The tensile strain in this case is defined as the ratio of the change in length ∆𝐿 to the original length 𝐿!. ∆𝐿 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝜀 = 𝐿! We define Young’s 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 by a combination of these two ratios: 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝜎 𝐹 ⁄𝐴 Units: N/m2 Young’s 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑌 = = 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝜀 ∆𝐿⁄𝐿! Young’s 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 is used to characterize a rod or wire stressed under either tension or compression. 29  Stress – Strain Curve Stress versus strain curve is a straight line, for relatively small stresses, the bar returns to its initial length when the force is removed. The elastic limit of a substance is defined as the maximum stress that can be applied to the substance before it becomes permanently deformed and does not return to its initial length. It is possible to exceed the elastic limit of a substance by applying a sufficiently large stress. As the stress increases, however, the curve is no longer a straight line. When the stress exceeds the elastic limit, the object is permanently distorted and does not return to its original shape after the stress is removed. As the stress is increased even further, the material ultimately breaks. 30  Shear Modulus: Elasticity of Shape Another type of deformation occurs when an object is subjected to a force parallel to one of its faces while the opposite face is held fixed by another force. The stress in this case is called a shear stress. If the object is originally a rectangular block, a shear stress results in a shape whose cross-section is a parallelogram. To a first approximation (for small distortions), no change in volume occurs with this deformation. 31  Shear Modulus: Elasticity of Shape We define the shear stress as the ratio of the tangential force to the area A of the face being sheared. 𝐹 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝐴 The shear strain is defined as the ratio of ∆x is the horizontal distance that the sheared face moves to h is the height of the object. ∆𝑥 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 𝜀 = ℎ In terms of these quantities, the Shear Modulus is 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝐹 ⁄𝐴 Units: N/m2 𝑆ℎ𝑒𝑎𝑟 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑆 = = 𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛 ∆𝑥⁄ℎ 32  Bulk Modulus: Volume Elasticity Bulk modulus characterizes the response of an object to changes in a force of uniform magnitude applied perpendicularly over the entire surface of the object. Such a uniform distribution of forces occurs when an object is immersed in a fluid. An object subject to this type of deformation undergoes a change in volume but no change in shape. 33  Bulk Modulus: Volume Elasticity The Volume Stress is defined as the ratio of the magnitude of the total force F exerted on a surface to the area A of the surface. The quantity 𝑃 = 𝐹 ⁄𝐴 is called pressure, if the pressure on an object changes by an amount ∆𝑃 = ∆𝐹 ⁄𝐴, the object experiences a volume change ∆𝑉. ∆𝐹 𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝜎 = = ∆𝑃 𝐴 The Volume Strain is equal to the change in volume ∆𝑉 divided by the initial volume 𝑉!. ∆𝑉 𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝜀 = 𝑉! Therefore we can characterize a volume (“bulk”) compression in terms of the Bulk Modulus, which is defined as 𝑉𝑜𝑙𝑢𝑚𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 ∆ 𝐹 ⁄𝐴 ∆𝑃 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐵 = =− =− 𝑉𝑜𝑙𝑢𝑚𝑒 𝑆𝑡𝑟𝑎𝑖𝑛 ⁄ ∆𝑉 𝑉! ∆𝑉 ⁄𝑉! 34  Bulk Modulus: Volume Elasticity A negative sign is inserted in this defining equation so that B is a positive number. This manoeuvre is necessary because an increase in pressure (positive ∆P) causes a decrease in volume (negative ∆V) and vice versa. If you look up such values in a different source, you may find the reciprocal of the Bulk Modulus listed. The reciprocal of the Bulk Modulus is called the compressibility of the material. 1 𝐾= 𝐵 35 36  Bulk Modulus: Volume Elasticity Notice from Table that both solids and liquids have a Bulk Modulus. No Shear Modulus and no Young’s Modulus are given for liquids, however, because a liquid does not sustain a shearing stress or a tensile stress. If a shearing force or a tensile force is applied to a liquid, the liquid simply flows in response. Consequently, an important characteristic of a fluid from the viewpoint of fluid mechanics is its compressibility. In other words, a fluid increases its pressure against compression, trying to retain its original volume. This characteristic is called compressibility. Another characteristic is its viscosity, a fluid shows resistance whenever two layers slide over each other. This characteristic is called Viscosity. Whereas a solid shows its elasticity in tension, compression or shearing stress, a fluid does so only for compression. 37  Ductile and Brittle Materials A material is said to be ductile A brittle material breaks soon if it can be stresses beyond its after the elastic limit is reached. elastic limit without breaking. 38 1- Assume that Young’s modulus is 1.5x1010 N/m2 for bone and that the bone will fracture if stress greater than 1.5x108 N/m2 is imposed on it. (a) What is the maximum force that can be exerted on the femur bone in the leg if it has a minimum effective diameter of 2.50 cm? (b) If this much force is applied compressively, by how much does the 25.0-cm-long bone shorten? Solution (a) 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝐹 ⁄𝐴 𝐹 = 𝑠𝑡𝑟𝑒𝑠𝑠×𝐴 d= 2.5 cm r= 1.25 cm 𝐴 = 𝜋𝑟 % 𝐹 = 1.5×10R 𝑁⁄𝑚% ×𝜋 1.25×10S%𝑚 % 𝐹 = 73631.08𝑁 From the definition of Young’s modulus: 𝑠𝑡𝑟𝑒𝑠𝑠 𝐹 ⁄𝐴 𝑌= = 𝑠𝑡𝑟𝑎𝑖𝑛 ∆𝐿⁄𝐿! 𝐹 ⁄𝐴 (73631.08𝑁)⁄(𝜋 1.25×10S%𝑚 %) ∆𝐿 = = $H % S% = 0.0025𝑚 𝑌⁄𝐿! (1.5×10 𝑁/𝑚 )⁄(25×10 𝑚)