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Chapter 7: Energy of a System Chapter 7 Goals: • To learn what constitutes a system of bodies • To define the work done by a force when a body moves • To generalize this definition to higher dimensions and/or position-varying forces and/or curvy paths • To define the kinetic energy and its changes, as related to the net force • To understand the circumstances in which a force possesses a potential energy, and to express it • To introduce the notion of mechanical energy conservation Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What is a System? • may be a single body, and all the forces that act on the body • they are dubbed external forces • may be a collection of bodies, and all the third-law pairs that act on the collection, and all external forces that act on the collection • the part of the universe that is exterior to the boundary is the environment • external forces get in through the boundary • other stuff cross the boundary: mass, heat, charge… • a system is often a rigid (or not so rigid) extended body • The universe = system environment Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Four increasingly complex scenarios involving the combination of force and motion • 1d constant force, motion in 1d • 1d varying force, motion in 1d • vector constant force, vector displacement • finally, a varying vector force with motion along a curvy path • sometimes, the path may be regarded as merely the displacement: but only if the force is conservative!! • a conservative force is one for which the work is pathindependent!! • note that work is done in a process • if the force is the net force, cool things can be said: kinetic energy can be defined and utilized!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Scenario the first • 1d constant force, motion in 1d • Work in moving from xi to xf under the influence of F: W [i f ] : Fx "force times distance" Scenario the Second • consider a force that varies with a body’s position F(x) • if the body moves a small displacement dx, so small that F(x) doesn’t change much, in the same (or opposite) direction as the force, the small bit of work done by the force is dW := F(x) dx • this is of course the (signed) area of the thin strip of width dx and height F(x) on a graph of F versus x • for a non-small displacement (from xi to xf), the work done by the force is the integral of dW (area under F(x)): Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Work as an integral in 1d xf W [ xi x f ] : F ( x) dx xi • the work is the signed area unde r the graph of F(x) • could be positive, negative or zero • note that if F(x) is a constant (first scenario) then ‘force times distance’ is the area of a simple rectangle Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The definition of Kinetic Energy, and the Work-Energy Theorem xf • if it’s the net force we get Wnet [i f ] : Fnet ( x) dx xi • Since Fnet = ma = m dv/dt we have (watch closely!!) xf Wnet [i f ] : xi xf Fnet ( x) dx m xi xf xf 2 f xi xi i dv dx mv dx m dv m v dv dt dt 2 1 2 • Define Kinetic Energy of a body K : mv 2 • Thus we have the Work-Energy Theorem: the work done on a body by the net force that acts in some process is equal to the change in kinetic energy of the body during that process, or Wnet = K:= Kf - Ki Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Units of Work and Energy • [W] = N-m = kg-m2/s2 = Joule (J) • If a force of 1 N acts on a body, and the body moves in the direction of the force a distance of 1 m, the force does 1 J of work (pretty small…) • If a force of 1 dyne (= 1 g-cm/s2) acts, and the body moves1 cm, the force does 1 erg of work (really small…) • If a force of 1 lb… moves 1 ft… 1 foot-pound of work • 1 calorie (cal) = 4.186 J [mechanical equivalent of heat] • 1 British Thermal Unit (BTU) = amount of heat needed to raise 1 lb of water 1 °F = 1055 J • 1 kcal = 1 Cal = 1 food calorie • 1 electron-Volt (ev) = 1.602x10-19 J • 1 kilowatt-hr = another common energy unit… Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Scenario the Third • consider a constant vector force that acts on a body, F • the force does not vary with a body’s position r • let the body move a vector displacement r:=rf –ri • done by the force is W[if] := F •r = |F||r| cos q where q is the angle between F and r when arranged tail-to-tail • thus the work is a scalar, no matter what • the more the motion is ‘with’ the force, the bigger the work done by the force • if the motion is perpendicular to the force, NO work! • if the motion is ‘against’ the force, then W < 0 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Reminder of scalar product facts • dot product can be thought of as magnitude of one, times the component of the other along the line of the first • in cartesian coordinates, A•B = AxBx + AyBy + AzBz • for any vector A , A•A = A2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Here the forces are constant. Which ones do positive work? Which negative work? Which none? W F r ' force dotted into displacement' Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Scenario the Fourth • consider a vector force that acts on a body, F(r) • the force varies with body’s position r as it moves along its path, so the answer may well depend on the path • if the body moves a small vector displacement dr, so small that F doesn’t change much, the small bit of work done by the force is dW := F ∙dr = |F|| dr| cos q • for a non-small displacement (from ri to rB), the work done by the force is the integral of dW r f W [i f ] : F(r ) dr r i • this is known as a line integral • it has to ‘follow’ the path • how can this be thought of as an area under a graph? • it is, in 3d, the sum of 3 areas! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Visualizing the work line integral Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Re-expressing this complex idea • write the force, and the arbitrary small diplacement, using components/unit vectors F Fx ˆi Fy ˆj Fz kˆ and dr dx ˆi dy ˆj dz kˆ • using the fact that the unit vectors are perpendicular, only three of the nine possible terms are non-zero: r W [i f ] : f r f r f r f r f dW F(r) dr Fx dx Fy dy Fz dz r i r i r i r i r i • in a sense, there is ‘x-work’ and ‘y-work’ and ‘z-work’ • see calculus III for the details of line integrals • again, the result may well be path-dependent… but in special cases it is not: conservative forces Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The (~same ) definition of Kinetic Energy, and the Work-Energy Theorem survives!! r f • If the force is the net force we get • Since Fnet = ma = m dv/dt we have Wnet rB rf rf ri ri rA Wnet : Fnet (r ) dr r i dv dr m dr m dv m v dv dt dt r f m v x dvx 2 more terms, for y and z ri m 2 2 v xf v 2yf v zf2 v xi v 2yi v zi2 2 m m m 2 v f v f v i v i K where K v 2 2 2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Is it possible to ‘unintegrate’ the work, to obtain the force? • YES it is possible but only if the force is conservative • in this situation, we DEFINE the work done by the force as the negative of the change in a different type of energy: the potential energy (of the force) • so the kinetic energy is owned by the body’s mass and speed, whereas the potential energy is owned by ‘potential energy field’, by virtue of the body’s position • alternative definition: given a force F(x), and a second force that ‘you’ exert that exactly opposes it, so Fyou = – F(x), the change in the potential energy of F(x) is equal to the work you do in overcoming (barely) F(x), as the motion proceeds (backward, of course) • there’s a bit of a pun here as you’ll see.. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. How is this expressed mathematically? • Assume that F(x) is conservative so that the potential energy exists; call it U(x) x x • ha ha ha f f U : U ( x f ) U ( xi ) : F ( x) dx Fyou ( x) dx x i x i • we see that since U is basically the integral of F, F must be the derivative of U (some details need to be clearer)! dU ( x) F ( x) : where x is a position coordinate ( y, z...) dx • in three dimensions the vector force is expressed as the gradient of the scalar potential energy function Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What are some potential energies in nature? • example: the force of gravity Fg = − mg • the 1d coordinate is y, and +y is ↑ yf yf yi yi U : U ( y f ) U ( yi ) : (mg ) dy mg dy mg ( y f yi ) • we assume initial y to be 0, and take U(yi) = 0 too • we assume final y to be just y, so we get U(y) • U := Ug(y) = mgy • so we conclude that the potential energy of an object a height y above the ‘ground’ is mass x g x height • check: what is (minus) this function’s space derivative? Fg ( y) dU g ( y) dy d (mgy) dy mg mg dy dy Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A New Force That’s Conservative: The Spring Force There does NOT need to be a block, yet!!! Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Wonders of the Spring Force • it is much like tension or compression, depending whether the spring is stretched longer than equilibrium position, or squeezed shorter • there is a very elegant expression for the force exerted BY the spring: Hooke’s Law F = – k (x – x0) • here, k is the spring constant • x is the location of the movable end of the spring • x0 is a constant too: the location of the movable end when the spring is neither stretched nor squeezed • often, one takes x0 to be zero, and we will take the potential energy of this force to be zero for x = x0 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A graph of Hooke’s law F(x) Stiff spring F exerted BY spring Limp spring x0 • Slope of the graph is –k • [k] = N/m Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. x What kind of acceleration occurs to a mass I? • At this instant, the spring is squeezed (x < 0) so its force is to the right (F > 0) • Therefore, acceleration of mass is to the right • If it is moving to the right, it will be speeding up and fly right through equilibrium, since a = 0 there • If it is moving to the left, it will be slowing down and eventually stop, since a is growing and opposed to the motion {show Active Figure AF_1502} Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What kind of acceleration occurs to a mass II? • The only horizontal force on the mass is the spring (ignore up/down forces) • Fnet = – kx = ma a = – kx/m • So the acceleration is in the opposite direction to the position, and proportional to it!! d 2x k x 0 a differential equation for x(t ) 2 m dt {show Active Figure AF_1501} Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. What would the potential energy be? xf xf xi xi U s : U s ( x f ) U s ( xi ) : (kx) dx k x dx 12 k x 2f xi2 • so this is the change in the potential energy if the spring end moves from xi to xf • has nothing to do with the mass except for where it is • we adjust the thinking now, and integrate from the reference position x0 (which we have taken to be zero: the equilibrium spring position) to the present position x: U s : U s ( x) U s ( x0 ) 12 k x 2 x02 12 k x 2 0 2 12 kx2 U s ( x) 12 kx2 where we take x0 0 and U s ( x0 ) 0 U s ( x) 12 k ( x x0 )2 if x0 0 and so U s ( x0 ) 0 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Recovering the Force dU s ( x) Fs ( x) dx d 1 k ( x x0 ) 2 k ( x x0 ) dx 2 The Physics Resides in V(x), if it exists • potential energies that depend only on position are conservative • the classic non-conservative force is friction • it depends on the state of motion, not merely on location Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Potential energy example: Spring • Checking the integral as an area • For simplicity, take x0 = 0 • The end of the spring moves from xmax < 0 to 0 • The force of the spring is positive, and the area under the curve is too • Area = ½ (base)(height) = ½ (–xmax)((– k xmax) 1 2 Area kxmax 2 1 2 So U kxmax 2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Potential energy functions U(x) dU ( x) • What information is encoded in U(x)? F ( x) dx • for x > 0, slope of U is positive, so F is to left • for x < 0, slope of U is negative, so F is to right • A restoring force • for x = 0, slope of U is zero, so F is zero: stable equilibrium point • turning points at ± xmax, where E = U K = 0 • unstable equilibrium point: object tends to ‘fly away’ Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Energetics of Simple Harmonic Motion x(t ) A sin t [ x0 0] 1 2 A2 U kx sin 2 t 2 2 sin and sin2 v(t ) A cost 2 1 2 m 2 A2 kA K mv cos 2 t cos 2 t 2 2 2 2 kA2 kA E K U sin 2 t cos 2 t constant!! 2 2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.