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

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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 an isolated system any more. conservative force, and the force that a spring exerts on 3 any object attached to the spring is another example. 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 1 2 2 energy will become 0 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 ∴=y =1.84 m 2 2 „ ½ mvf + mgy = mgh ()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). to a maximum height of 20 m above the position of the projectile before firing. Neglecting all Answer: resistive forces, determine the spring constant k. ½ kx2 = ½ mv 2 + mgx Answer: elastic potential energy B B ½ x (953 N/m) (0.12 m)2 2 gravitational potential energy ½k kx = mgh 2 = ½ x (0.035 kg)vB + 2 = 2mgh / x (0.035 kg)(9.8 m/s2)(0.12 m) = 2 (0.035 kg)(9.8 m/s2)(20 m) / (0.12 m)2

= 953 N/m vB = 19.7 m/s 21 22

Mechanical Energy and Nonconservative Forces Nonconservative Forces

„ A nonconservative force does not „ In general, if friction is acting in a satisfy the conditions of conservative system:

forces „ ΔEmech = ΔK + ΔU = -ƒkd „ Nonconservative forces acting in a „ Difference in signs is due to the loss and system cause a change in the gain in the energy mechanical energy of the system „ ΔU is the change in all forms of potential energy Recall that conservative forces ha „ If friction is zero, this equation becomes •The work done by a conservative fo the same as Conservation of Mechanical any two points is Energy, E = 0, i.e., no change in •The work done by a conservative fo Δ mech any closed path isindependent zero. ve these two equivalent properties: 23 mechanical energy. 24 rce on a particle moving between of the path taken by the particle. rce on a particle moving through Nonconservative Forces, Nonconservative Forces, cont Example 1 (Slide)

„ The work done against friction is greater along the brown path than ΔEmech = ΔK + ΔU

along the blue path ΔEmech =(Kf – Ki) + (Uf – Ui)

„ Because the work done ΔEmech = (Kf + Uf) – (Ki + Ui) depends on the path, 2 ΔEmech = ½ mvf – mgh friction is a = -ƒ d nonconservative force k

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We will first analyze the sequence of arrivals of these Example. Mary and John are at a water park. There g sin Ө are two water slides that start with the same height and metal balls on the tracks. Which metal ball will reach end at the same height. Slide A has more gradual slope the bottom first? than slide B. John likes slide B better and he says he g Ө C reaches at a faster speed with slide B when touching D the water because he notes that he got to the bottom level in less time on slide B than on slide A as measured A B with his stop watch. Mary who does not carry a stop watch says she touches the water with the same speed g sin Ө on either slide. Who is correct and why? Both slides

have negligible friction. (demo 5 tracks) g A Ө Answer: Mary is correct. The velocity when they touch the B water is the same at either slide because the vertical C D distance of decent is the same. mghmv= 1 2. 2 John is partially correct. As the acceleration along the slide (g sin Ө) is larger on B, the velocity is increased at a faster B rate on slide B during the earlier duration so it takes lesser time (or is faster) to get to the bottom on slide B, but the A 27 final speeds are the same on both slides. 28 Example. A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s. The block comes to rest after traveling 3.00 m Nonconservative Forces, along the plane, which is inclined at an angle of 30.0° to the horizontal. For this motion determine Example 2 (Spring-Mass) (a) the change in the block's kinetic energy, 30° (b) the change in the potential energy of the „ Without friction, the block-Earth system, and (c) the friction force 30° exerted on the block (assumed to be constant). mg energy continues to be (d) What is the coefficient of kinetic friction? transformed between 1122 2 kinetic and elastic Answer: (a) Δ=Kmvv( −) =− mv =−160 J 22fi i potential energies and (b) Δ=Umg(3.00 msi) n30.0 °= 73.5 J the total energy (c) The mechanical energy converted to heat due to friction is remains the same 160 J – 73.5 J = 86.5 J. As W = f x d, f = W /d, so „ If friction is present, the 86.5 J f==28.8 N energy decreases 3.00 m „ ΔEmech = -ƒkd (d) fnmg= μkk=°=μ cos30.0 28.8 N 28.8 N μ ==0.679 k 2 ()5.00 kg() 9.80 m s cos30.0° 29 30

Nonconservative Forces, Example 3 (Connected Blocks) Connected Blocks, cont

„ The system consists of the two blocks, the spring, and „ Block 2 undergoes a Earth change in gravitational potential energy „ Gravitational and potential „ The spring undergoes a energies are involved, and change in elastic potential friction is not 0 energy

„ The kinetic energy is zero if „ The coefficient of kinetic our initial and final friction can be measured configurations are at rest „ Why is the coefficient of static friction not used? 31 32 Example. A 1.00-kg object slides to the right on a (b) the speed v at the unstretched position surface having a coefficient of kinetic friction 0.250. The when the object is moving to the left object has a speed of vi = 3.00 m/s when it makes (figure 4) contact with a light spring that has a force constant of 50.0 N/m. The object comes to rest after the spring has Answer: been compressed a distance d. The object is then forced toward the left by the spring and continues to move in Now the spring is at the natural position so the elastic that direction beyond the spring’s unstretched position. potential energy in the spring is zero. The remaining Finally the object comes to rest a distance D to the left of the unstretched spring. Find (a) the distance of energy left for pushing the block to the left is the compression d, (b) the speed v at the unstretched initial kinetic energy less the workdone by friction position when the object is moving to the left, and (c) the where the block has travelled a distance of 2d. distance D where the object comes to rest. 2 Answer: X m (a) Initial amount of energy in the system is . = This energy is given to the elastic potential energy in the spring when it is compressed and the workdone by friction (figure 3), i.e., = +

= + f = μmg = 0.25 x 1 x 9.8 By the roots of quadratic = 2.45 N 2 equation ax +bx + c = 0 Æ 33 34

(c) the distance D where the object comes to rest.

Answer: Energy Diagrams and Stable As the spring is elastic, the potential energy gained and subsequently returned by the Equilibrium spring has a net 0 value (we assume that no heat is generated in the spring or it is negligible as the deformation is done only „ The x = 0 position is once). oneequilibrium of stable When the (detached) object comes to rest, the initial kinetic energy is transformed to the work „ Configurations of done by friction travelled with a distance stable equilibrium of ( d + d + D). correspond to those for which potential A ball in a valley is in = the state of stable energy U(x) is a equilibrium. minimum

35 36 Energy Diagrams and Unstable Equilibrium Neutral Equilibrium

„ F = 0 at x = 0, so the x „ Neutral equilibrium occurs in a particle is in equilibrium „ For any other value of configuration when U is constant over x, the particle moves some region away from the equilibrium position „ A small displacement from a position in „ This is an example of this region will produce neither unstable equilibrium restoring nor disrupting forces „ Configurations of A ball on top of a hill is unstable equilibrium in the state of unstable A ball on a flat ground correspond to those for equilibrium. is in the state of neutral which U(x) is a equilibrium. maximum 37 38

Example. A particle moves along a line where the (a) Identify each potential energy of its system equilibrium position for depends on its position r as this particle. Indicate graphed in Figure. In the limit whether each is a point of as r increases without bound, stable, unstable or U(r) approaches +1 J. neutral equilibrium. (a) Identify each equilibrium Answer: position for this particle. Indicate whether each is a point of stable, unstable or neutral There is an equilibrium point wherever tangent to the graph equilibrium. of potential energy is horizontal:

At r = 1.5 mm and 3.2 mm, the equilibrium is stable. Now suppose that the system (PE + KE) has energy −3 J. Determine (b) the range of positions where the particle can At r = 2.3 mm, the equilibrium is unstable. be found, (c) its maximum kinetic energy, (d) the location where it has maximum kinetic energy. A particle moving out toward r →∞ approaches neutral 39 equilibrium. 40 Now suppose that the system has energy −3 J. Now suppose that the system has energy −3 J. Determine (b) the range of (d) Find the location where it has positions where the particle can maximum kinetic energy be found

Answer: If the system energy is Answer: –3 J and the system is (c) Find its maximum isolated, its potential kinetic energy Kinetic energy is a maximum energy must be less Answer: when the potential energy is a than or equal to –3 J. KU+= E minimum, at r= 1.5 mm Thus, the particle’s =− =− −−( )= position is limited to KEUmax min 3.0 J5.6 J 0.6 mm≤≤r 3.6 mmKEUmax=− min =−3.0 J5. −−()6 J2. = 6 J 41 42

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