Physics I course – Winter 2021 Dr. Yehia Eissa Lecturer of Physics, Faculty of Engineering German International University (GIU) 1 Learning outcomes Objective: The major objectives of this course are for students to learn the fundamental principles of classical mechanics, to develop solid and systematic problem solving skills, and to lay the foundations for further studies in physics, physical sciences, and engineering. Topics: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. Physics and measurement Motion in one dimension Vectors Motion in two dimensions The laws of motion Energy and energy transfer Potential energy Linear momentum and collisions Rotation of a rigid object about a fixed axis Angular momentum Static equilibrium and elasticity Fluid mechanics Introduction to oscillatory motion 2 References 3 Chapters 7 & 8: Energy Dec. 18, 2021 4 Doing work, transferring energy • The force exerted by the weightlifter transfers energy from her to the weights. We know that the weights have gained energy because, when the athlete releases them, they come crashing down to the ground. • ‘Doing work’ is a way of transferring energy from one object to another. 5 Work The work, W, done on a system by an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Dr of the displacement of the point of application of the force, and cos q, where q is the angle between the force and the displacement vectors. The meaning of the term work is distinctly different in physics than in everyday meaning. Work is done by some part of the environment that is interacting directly with the system. Work is done on the system. 6 Work, cont. W = F Dr cos q The displacement is that of the point of application of the force. A force does no work on the object if the force does not move through a displacement. The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application. 7 Work Example The normal force and the gravitational force do no work on the object. cos q = cos 90° = 0 The force F is the only force that does work on the object. 8 More About Work The sign of the work depends on the direction of the force relative to the displacement. Work is positive when projection of as the displacement. F onto Dr is in the same direction Work is negative when the projection is in the opposite direction. The work done by a force can be calculated, but that force is not necessarily the cause of the displacement. Work is a scalar quantity. The unit of work is a joule (J) 1 joule = 1 newton . 1 meter = kg ∙ m² / s² J=N·m The joule is defined as the amount of work done when a force of 1 newton moves a distance of 1 meter in the direction of the force. 9 Work Is An Energy Transfer This is important for a system approach to solving a problem. If the work is done on a system and it is positive, energy is transferred to the system. If the work done on the system is negative, energy is transferred from the system. If a system interacts with its environment, this interaction can be described as a transfer of energy across the system boundary. This will result in a change in the amount of energy stored in the system. 10 Example 7.1 • A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F = 50.0 N at an angle of 30.0° with the horizontal. • Calculate the work done by the force on the vacuum cleaner as the vacuum cleaner is displaced 3.00 m to the right. 11 Example 7.1 – Solution 12 Scalar Product of Two Vectors The scalar product of two vectors is written as A B . It is also called the dot product. A B A B cos q q is the angle between A and B Applied to work, this means W F Dr cosq F Dr 13 Scalar Product, cont The scalar product is commutative. A B B A The scalar product obeys the distributive law of multiplication. A B C A B A C 14 Dot Products of Unit Vectors ˆi ˆi ˆj ˆj kˆ kˆ 1 ˆi ˆj ˆi kˆ ˆj kˆ 0 Using component form with vectors: A Ax ˆi Ay ˆj Azkˆ B Bx ˆi By ˆj Bzkˆ A B Ax Bx Ay By Az Bz In the special case where A B; A A Ax2 Ay2 Az2 A2 15 Example 7.2 16 Example 7.2 – Solution 17 Example 7.3 18 Example 7.3 – Solution 19 Work Done by a Varying Force To use W = F Δr cos θ, the force must be constant, so the equation cannot be used to calculate the work done by a varying force. Assume that during a very small displacement, Dx, F is constant. For that displacement, W ~ F Dx For all of the intervals, xf W Fx Dx xi 20 Work Done by a Varying Force, cont. Let the size of the small displacements approach zero . Since lim Dx 0 xf F Dx x xi xf xi Fx dx Therefore, xf W Fx dx xi The work done is equal to the area under the curve between xi and xf. 21 Work Done By Multiple Forces If more than one force acts on a system and the system can be modeled as a particle, the total work done on the system is the work done by the net force. W W ext xf xi F dx x In the general case of a net force whose magnitude and direction may vary. W W ext xf xi Fdr The subscript “ext” indicates the work is done by an external agent on the system. 22 Example 7.4 • A force acting on a particle varies with x, as shown in the figure. • Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m. 23 Example 7.4 – Solution 24 Example 7.5 25 Example 7.5 – Solution 26 Work Done By A Spring A model of a common physical system for which the force varies with position. The block is on a horizontal, frictionless surface. Observe the motion of the block with various values of the spring constant. 27 Spring Force (Hooke’s Law) The force exerted by the spring is Fs = - kx x is the position of the block with respect to the equilibrium position (x = 0). k is called the spring constant or force constant and measures the stiffness of the spring. k measures the stiffness of the spring. This is called Hooke’s Law. 28 Hooke’s Law, final The force exerted by the spring is always directed opposite to the displacement from equilibrium. The spring force is sometimes called the restoring force. If the block is released it will oscillate back and forth between –x and x. 29 Work Done by a Spring Identify the block as the system. Calculate the work as the block moves from xi = - xmax to xf = 0. 0 𝑊𝑠 = න 𝐹𝑆 𝑑𝑥 = න −𝑥𝑚𝑎𝑥 1 2 −𝑘𝑥 𝑑𝑥 = 𝑘𝑥𝑚𝑎𝑥 2 The net work done as the block moves from -xmax to xmax is zero 30 Work Done by a Spring, cont. Assume the block undergoes an arbitrary displacement from x = xi to x = xf. The work done by the spring on the block is Ws xf xi kx dx 1 2 1 2 kxi kxf 2 2 If the motion ends where it begins, W = 0 31 Example 7.6 • A common technique used to measure the force constant of a spring is demonstrated by the setup in the figure. • The spring is hung vertically, and an object of mass m is attached to its lower end. Under the action of the “load” mg, the spring stretches a distance d from its equilibrium position. • If a spring is stretched 2.0 cm by a suspended object having a mass of 0.55 kg, what is the force constant of the spring? • How much work is done by the spring as it stretches through this distance? 32 Example 7.6 – Solution 33 Kinetic Energy One possible result of work acting as an influence on a system is that the system changes its speed. The system could possess kinetic energy. Kinetic Energy is the energy of a particle due to its motion. K = ½ mv2 K is the kinetic energy m is the mass of the particle v is the speed of the particle A change in kinetic energy is one possible result of doing work to transfer energy into a system. 34 Kinetic Energy, cont Calculating the work: Wext xf xi F dx xf xi vf Wext mv dv vi Wext Wext 35 1 1 2 mv f mv i2 2 2 K f K i DK ma dx Work-Kinetic Energy Theorem The Work-Kinetic Energy Theorem states Wext = Kf – Ki = ΔK When work is done on a system and the only change in the system is in its speed, the net work done on the system equals the change in kinetic energy of the system. The speed of the system increases if the work done on it is positive. The speed of the system decreases if the net work is negative. The work-kinetic energy theorem is not valid if other changes (besides its speed) occur in the system or if there are other interactions with the environment besides work. The work-kinetic energy theorem applies to the speed of the system, not its velocity. 36 Work-Kinetic Energy Theorem – Example The block is the system and three external forces act on it. The normal and gravitational forces do no work since they are perpendicular to the direction of the displacement. Wext = DK = ½ mvf2 – 0 The answer could be checked by modeling the block as a particle and using the kinematic equations. 37 Work-Kinetic Energy Theorem – Example • A 6.0-kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N. • Find the speed of the block after it has moved 3.0 m. 38 Work-Kinetic Energy Theorem – Example 39 Any questions? 40