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Kinetic Energy, Work, Power, and Potential Energy 8.01 W05D1 Today’s Reading Assignment: W05D1 Young and Freedman: 6.1-6.4 Math Review Module: Scalar Product Kinetic Energy • Scalar quantity (reference frame dependent) 1 2 K mv 0 2 • SI unit is joule: 1J 1kg m2 /s 2 • Change in kinetic energy: 1 2 1 2 1 1 2 2 2 2 2 K mv f mv0 m(vx, f v y, f vz, f ) m(vx,0 v 2y,0 vz,0 )0 2 2 2 2 Concept Question: Work and Kinetic Energy Compared to the amount of energy required to accelerate a car from rest to 10 mph (miles per hour), the amount of energy required to accelerate the same car from 10 mph to 20 mph is (1) the same (2) twice as much (3) three times as much (4) four times as much (5) unsure. Work Done by a Constant Force for One Dimensional Motion Definition: The work W done by a constant force with an x-component, Fx, in displacing an object by x is equal to the xcomponent of the force times the displacement: W Fx x Concept Question: Pushing against a wall The work done by the contact force of the wall on the person as the person moves away from the wall is 1. positive. 2. negative. 3. zero. 4. impossible to determine from information given in question and the figure. Concept Question: Work and Walking When a person walks, the force of friction between the floor and the person's feet accelerates the person forward. The work done by the friction force is 1. positive. 2. negative. 3. zero. Pushing a Stalled Car Table Problem: Work Done by Gravity Near the Surface of the Earth Consider an object of mass near the surface of the earth falling directly towards the center of the earth. The gravitational force between the object and the earth is nearly constant. Suppose the object starts from an initial point that is a distance y0 from the surface of the earth and moves to a final point a distance yf from the surface of the earth. How much work does the gravitational force do on the object as it falls? Work done by Non-Constant Force: One Dimensional Motion (Infinitesimal) work is a scalar Wi (Fx )i xi Add up these scalar quantities to get the total work as area under graph of Fx vs x : iN iN i 1 i 1 W Wi ( Fx )i xi As N and xi 0 W lim iN ( F ) x N xi 0 i 1 x i i x x f x x0 Fx dx Concept Question: Work due to Variable Force A particle starts from rest at x = 0 and moves to x = L under the action of a variable force F(x), which is shown in the figure. What is the particle's kinetic energy at x=L/2 and at x=L? (1) (Fmax)(L/2), (Fmax)(L) (2) (Fmax)(L/4), 0 (3) (Fmax)(L), 0 (4) (Fmax)(L/4), (Fmax)(L/2) (5) (Fmax)(L/2), (Fmax)(L/4) Table Problem: Work Done by the Spring Force Connect one end of a spring of length l0 with spring constant k to an object resting on a smooth table and fix the other end of the spring to a wall. Stretch the spring until it has length l and release the object. How much work does the spring do on the object as a function of x = l - l0, the distance the spring has been stretched or compressed? Worked Example: Work Done by Several Forces A block of mass m slides along a horizontal table with speed v0. At x = 0 it hits a spring with spring constant k and begins to experience a friction force. The coefficient of kinetic friction is given by m. How far did the spring compress when the block first momentarily comes to rest? Recall integration formula for acceleration with respect to time The x-component of the acceleration of an object is the derivative of the x-component of the velocity dv x ax dt Therefore the integral of x-component of the acceleration with respect to time, is the xcomponent of the velocity tf t0 ax dt tf t0 dvx dt dt vx , f vx ,0 dvx vx, f vx,0 Integration formula for acceleration with respect to displacement The integral of x-component of the acceleration with respect to the displacement of an object, is given by xf x f dv xf vx , f dx x x0 ax dx x0 dt dx x0 dvx dt vx ,0 vx dvx 1 2 2 a dx d (1 / 2)v (v v ) x x, f x,0 x0 x vx ,0 2 Multiply both sides by the mass of the object giving integration formula xf xf x0 max dx vx , f vx , f vx ,0 d (1 / 2)vx 2 2 1 2 1 2 mvx, f mvx,0 K 2 2 Work-Kinetic Energy Theorem One Dimensional Motion Substitute Newton’s Second Law (in one dimension) Fx max in definition of work integral which then becomes W xf x0 Fx dx xf x0 max dx Apply integration formula to get work-kinetic energy theorem W xf x0 Fx dx xf x0 max dx K Concept Question: WorkEnergy An object is dropped to the earth from a height of 10m. Which of the following sketches best represent the kinetic energy of the object as it approaches the earth (neglect friction)? 1. 2. 3. 4. 5. a b c d e Concept Question Two objects are pushed on a frictionless surface from a starting line to a finish line with equal constant forces. One object is four times as massive as the other. Both objects are initially at rest. Which of the following statements is true when the objects reach the finish line? 1. The kinetic energies of the two objects are equal. 2. Object of mass 4m has the greater kinetic energy. 3. Object of mass m has the greater kinetic energy. 4. Not information is given to decide. Worked Example: Work-Energy Theorem for Inverse Square Gravitational Force Consider a magnetic rail gun that shoots an object of mass m radially away from the surface of the earth (mass me). When the object leaves the rail gun it is at a distance ri from the center of the earth moving with speed vi . What speed of the object as a function of distance from the center of the earth? Power • The average power of an applied force is the rate of doing work W Fapplied, x x P Fapplied,x vx t t • SI units of power: Watts 1W 1J/s 1kg m /s 2 • Instantaneous power 3 W x P lim Fapplied,x lim Fapplied,x vx t0 t t0 t Work and the Dot Product Dot Product A scalar quantity Magnitude: A B A B cos The dot product can be positive, zero, or negative Two types of projections: the dot product is the parallel component of one vector with respect to the second vector times the magnitude of the second vector A B A (cos ) B A B A B A (cos ) B A B Dot Product of Unit Vectors in Cartesian Coordinates For unit vectors We have φ i, φ j ˆi ˆi | ˆi || ˆi | cos(0) 1 ˆi ˆj | ˆi || ˆj |cos( /2) 0 Generally: φ iφ i φ j φ j kφ kφ 1 φ iφ j φ i kφ φ j kφ 0 Example: Dot Product in Cartesian Coordinates r φ A Ax φ i Ay φ j Az k, r B Bx φ i By φ j Bz kφ Then r r φ φ (B φ φ A B ( Ax φ i Ay φ j Az k) i B j B k) x y z Ax φ i Bx φ i Ax φ i By φ j Ax φ i Bz kφ Ay φ j Bx φ i Ay φ j By φ j Ay φ j Bz kφ Az kφ Bx φ i Az kφ By φ j Az kφ Bz kφ Ax Bx Ay By Az Bz Kinetic Energy and Dot Product Velocity r v vx φ i vyφ j vz kφ Kinetic Energy: 1 r r 1 K m( v v) m(vx2 v 2y vz2 ) 0 2 2 Change in kinetic energy: 1 2 1 2 1 r r 1 r r K mv f mv0 m( v f v f ) m( v 0 v 0 ) 2 2 2 2 1 1 2 2 2 2 2 m(vx, f v y, f vz, f ) m(vx,0 v 2y,0 vz,0 ) 2 2 Work Done by a Constant Force Definition: Work The work done by a constant force F on an object is equal to the component of the force in the direction of the displacement times the magnitude of the displacement: W F r F r cos F cos r F r Note that the component of the force in the direction of the displacement can be positive, zero, or negative so the work may be positive, zero, or negative Concept Question: Work and Gravity 1 A ball is given an initial horizontal velocity and allowed to fall under the influence of gravity near the surface of the earth, as shown below. The work done by the force of gravity on the ball is: (1) positive (2) zero (3) negative Worked Example: Work Done by a Constant Force in Two Dimensions Let the force exerted on an object be F F ˆi F ˆj x y Fx F cos Fy F sin Displacement: r x ˆi W F r F x cos ( Fx ˆi Fy ˆj) (x ˆi ) Fx x Table Problem: Work Constant Forces and Dot Product An object of mass m, starting from rest, slides down an inclined plane of length s. The plane is inclined by an angle of θ to the ground. The coefficient of kinetic friction is μ. a) Use the dot product definition of work to calculate the work done by the normal force, the gravitational force, and the friction force as the object displaces a distance s down the inclined plane. b) For each force, is the work done by the force positive or negative? c) What is the sum of the work done by the three forces? Is this positive or negative? Concept Question: Work and inverse square gravity A comet is speeding along a hyperbolic orbit toward the Sun. While the comet is moving away from the Sun, the work done by the Sun on the comet is: (1) positive (2) zero (3) negative Work Done Along an Arbitrary Path Work done by force for small displacement r r Wi Fi ri Work done by force along path from A to B r r lim Fi ri i N WAB N r ri 0 i1 B A r r F dr Work-Energy Theorem in Three-Dimensions As you will show in the problem set, the one dimensional work-kinetic energy theorem generalizes to three dimensions r B r r B r r B d vr r B r dr r r F d r ma d r m dr mdv mdv v dt dt A A A A A B WAB 1 2 1 2 mv B mv A K B K A 2 2 K Work: Path Dependent Line Integral Work done by force along path from A to B r r lim Fi ri i N WAB N r ri 0 i1 B A r r F dr In order to calculate the line integral, in principle, requires a knowledge of the path. However we will consider an important class of forces in which the work line integral is independent of the path and only depends on the starting and end points Conservative Forces Definition: Conservative Force If the work done by a force in moving an object from point A to point B is independent of the path (1 or 2), B Wc Fc dr (path independent) A then the force is called a conservative force which we denote by Fc . Then the work done only depends on the location of the points A and B. Example: Gravitational Force Consider the motion of an object under the influence of a gravitational force near the surface of the earth The work done by gravity depends only on the change in the vertical position Wg Fg y mg y Potential Energy Difference Definition: Potential Energy Difference between the points A and B associated with a conservative force Fc is the negative of the work done by the conservative force in moving the body along any path connecting the points A and B. B U Fc dr Wc A Potential Energy Differnece: Constant Gravity Force: r r F mg F y φ j mg φ j Work: W Fy y mgy Potential Energy: U W mgy mg y f y0 Choice of Zero Point: Choose y0 0 and choose U ( y0 ) 0 . Then Potential Energy: U ( y) mgy Worked Example: Change in Potential Energy for Inverse Square Gravitational Force Consider an object of mass m1 moving towards the sun (mass m2). Initially the object is at a distance r0 from the center of the sun. The object moves to a final distance rf from the center of the sun. For the object-sun system, what is the change in potential during this motion? Worked Example Solution: Inverse Square Gravity Force: Work done: Fm1 ,m2 rf Gm1m2 rˆ 2 r rf Gm1m2 Gm1m2 W F dr dr 2 r r r0 r0 Potential Energy Change: Zero Point: Potential Energy Function rf r0 1 1 Gm1m2 r r 0 f 1 1 U W Gm1m2 r f r0 U (r0 ) 0 Gm1m2 U (r) = r Table Problem: Change in Potential Energy Spring Force Connect one end of a spring of length l0 with spring constant k to an object resting on a smooth table and fix the other end of the spring to a wall. Stretch the spring until it has length l and release the object. Consider the objectspring as the system. When the spring returns to its equilibrium length what is the change in potential energy of the system? Potential Energy Difference: Spring Force Force: F Fx ˆi kxˆi xxf Work done: W x x0 1 kx dx k x 2f x02 2 Potential Energy Change: 1 U W k x 2f x02 2 Zero Point: U (x 0) 0 Potential Energy 1 2 U (x) kx 2 Work-Energy Theorem: Conservative Forces The work done by the force in moving an object from A to B is equal to the change in kinetic energy r r 1 2 1 2 W F d r mvB mv A K A 2 2 B When the only forces acting on the object are conservative forcesr r F Fc then the change in potential energy is U W Therefore U K Table Problem : Asteroid about Sun An asteroid of mass m is in a non-circular closed orbit about the sun. Initially it is a distance ri from the sun, with speed vi. What is the change in the kinetic energy of the asteroid when it is a distance is rf, from the sun? Next Reading Assignment: W05D2 Young and Freedman: 7.1-7.5,12.3 Experiment 3: Energy Transformations 44