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PC1221 Fundamentals of Physics I Ground Rules Isolated System Ground Rules PC1221 Fundamentals of Physics I Switch off your handphone and pager Lectures 15 and 16 Switch off your laptop computer and keep it No talking while lecture is going on No gossiping while the lecture is going on Potential Energy Raise your hand if you have question to ask Be on time for lecture Dr Tay Seng Chuan Be on time to come back from the recess break to continue the lecture Bring your lecturenotes to lecture 1 2 Conservative Isolated System Forces An isolated system is one for which there are no energy transfers across the boundary. The energy in such a system is Conservative forces have these two equivalent properties: conserved , i.e., at anytime the sum is a •The work done by a conservative force on a particle moving between any two points is independent of the constant but its form can change in part path taken by the particle. or in whole. E.g., a block sliding across a •The work done by a conservative force on a particle frictionless table is moving in an isolated moving through any closed path is zero. system. If there is friction on the table (rough surface), the block is not sliding in The gravitational force (free fall) is one example of a conservative force, and the force that a spring exerts on an isolated system any more. any object attached to the spring is another example. 3 4 Potential Energy System Example This system consists of Earth and Potential energy is the energy associated with the configuration of a a book system of objects that exert forces on Do work on the system by lifting each other. Eg, if you stand on top of Kent Ridge Hill, your potential energy is the book through Δy larger than what you have now in this The work done is mgyb – mgya lecture theatre at a lower ground. The energy storage mechanism is When conservative forces act within an isolated system, the kinetic energy gained called potential energy (or lost) by the system as its members change their relative positions is balanced Potential energy is always by an equal loss (or gain) in potential associated with a system of two or energy. more interacting objects This is Conservation of Mechanical Energy.Energy 5 6 Gravitational Potential Energy The gravitational potential energy depends only on the vertical height of Systems with Multiple Particles the object above Earth’s surface a In solving problems, you must choose We can extend our definition of a a reference configuration for which system to include multiple objects the gravitational potential energy is b set equal to some reference value, The force can be internal to the system normally zero The kinetic energy of the system is the The choice is arbitrary because you normally need the difference in algebraic sum of the kinetic energies of potential energy, which is independent of X the choice of reference configuration. the individual objects Therefore the choice of reference frame is not important. E.g., the distance Y between a and b is always the same regardless of its reference with respect to X or Y. 7 8 Conservation of Mechanical Conservation of Mechanical Energy Energy, example The mechanical energy of a system is the Look at the work done algebraic sum of the kinetic (K ) and potential by the book as it falls energies (Ug) in the system from some height to a Emech = K + Ug lower height The statement of Conservation of Mechanical Won book = ΔKbook Energy for an isolated system (An isolated Also, W = mgyb – mgya system is one for which there are no energy So, ΔK = -ΔU transfers across the boundary) is g Gain in Loss in K + U = K + U f f i i kinetic potential (i.e., final sum = initial sum) energy energy 9 10 Elastic Potential Energy Elastic Potential Energy is associated Elastic Potential Energy with a spring The force the spring exerts (on a block, for example) is Fs = - kx The work done by an external applied The elastic potential energy (U )stored in a spring is force on a spring-block system is 2 2 zero whenever the spring is not deformed (U = 0 W = ½ kxf –½kxi (x is measured from equilibrium position.) when x = 0) The work is equal to the difference between the initial and final values of an expression The energy is stored in the spring only when the spring is related to the configuration of the system stretched or compressed This expression is the elastic potential 2 The elastic potential energy is a maximum when energy: Us = ½ kx , where x is the distance from the natural position. the spring has reached its maximum extension or The elastic potential energy can be compression regarded as the energy stored in the 2 deformed spring The elastic potential energy (Us = ½ kx ) is always The stored potential energy can be positive because x2 will always be positive converted into kinetic energy 11 12 Conservation of Energy, Example. Dave Johnson, the bronze medalist at the 1992 Olympic decathlon in Barcelona, leaves the ground Example 1 (Drop a Ball) at the high jump with vertical velocity component 6.00 m/s. How far does his center of mass move up as he Initial conditions: makes the jump? Ei = Ki + Ui = mgh Answer: The ball is dropped from rest, so Ki = 0 The configuration for zero potential From leaving ground to the highest energy is the ground, i.e., when the point, += + ball hits the ground its potential KUii K f U f energy will become 0 1 2 2 mmy()6.00 m s+=+ 0 0 9.80 m s 2 () Conservation rules applied at some 2 point y above the ground gives ()6.00 m s ∴= = 2 y 1.84 m ½ mv + mgy = mgh 2 f ()29.()80 m s 13 14 Example. Three identical balls are thrown from the top of building all with the same Conservation of Energy, initial speed. The first is thrown horizontally, the second at some angle above the Example 2 (Pendulum) horizontal, and the third at some angle below the horizontal. Neglecting air As the pendulum swings, there is a continuous change resistance, rank the speeds of the balls at between potential and the instant each hits the ground kinetic energies At A, the energy is potential Answer: At B, all of the potential You might rank ball-2 with the highest speed because it travels to a energy at A is transformed higher point before it starts to fall. As the height will be higher so the into kinetic energy speed when ball-2 hits the ground will be larger. Let zero potential energy be at B In actual case, all the 3 speeds will be the same when the balls hit the At C, the kinetic energy has ground. This is because the initial positions of the balls are the same (so been transformed back into the potential energies are the same), and the initial speeds are the same potential energy (so the kinetic energies are the same). When the balls hit the ground, the total mechanical energy (potential + kinetic) for each ball is converted to kinetic energy. 15 16 Example. Jane, whose mass is What if we want the ball 50.0 kg, needs to swing across a to swing at least half a l: the length of the river (having width D) filled with circle about the pole? How? string man-eating crocodiles to save r: radius of the circle Tarzan from danger. She must Answer: l − 2r centered at the pole swing into a wind exerting cos Ө = Ө constant horizontal force F, on a l vine having length L and initially making an angle θ with the vertical. Taking D = 50.0 m, F = 110 N, L = 40.0 m, and θ = 50.0°. (a) With what minimum speed must Jane begin her swing in order to just make it to the other side? (b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume that Tarzan has a mass of 80.0 kg. 17 18 (a) With what minimum speed 0 potential level (b) Once the rescue is complete, 0 potential level must Jane begin her swing in Tarzan and Jane must swing back 4 0 across the river. With what order to just make it to the 110 N m other side? (reverse swing ball) minimum speed must they begin their swing? Assume that Tarzan Answer: Jane DL=+sinsiθ Lnφ has a mass of 80.0 kg. Answer: 50.0 m40.=°+0 msi( n50 sinφ) initial KE + initial PE + φ =°28.9 workdone by the wind force 50 m initial KE + initial PE – = final PE at destination workdone in overcoming the wind force 1 2 mvi +− mg() Lcosφθ ++=+− FD()= 1 0 mg ( L cos ) = final PE at destination 2 1 22 130 kgv +−°+ 130 kg 9.8 m s()() 40 m cos28.9 110 N50 m - FD = 2 i () 2 =−°130 kg() 9.8 m s() 40 mcos50 1 130 kgv24−× 4.46 10 J5500 + J3. =−×28 10 4 J 2 i 26340() J vi ==9.87 m s 19 130 kg 20 Conservation of Energy, Conservation of Energy, Example 3 (Spring Gun) Example 3 (Spring Gun) h The launching mechanism of a toy gun consists Find the speed of the projectile as it of a spring of unknown constant k. When the spring is compressed 0.12 m, the gun, when moves through the equilibrium position fired vertically, is able to launch a 35 g projectile of the spring (at XB = 0.12 m).
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