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PHSX 114, Monday, September 29, 2003 Reading for today: Chapter 6 (6-3 -- 6-7)Reading for today: Chapter 6 (6-3 -- 6-7) Reading for next lecture (Wed.):

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Presentation on theme: "PHSX 114, Monday, September 29, 2003 Reading for today: Chapter 6 (6-3 -- 6-7)Reading for today: Chapter 6 (6-3 -- 6-7) Reading for next lecture (Wed.):"— Presentation transcript:

1 PHSX 114, Monday, September 29, 2003 Reading for today: Chapter 6 (6-3 -- 6-7)Reading for today: Chapter 6 (6-3 -- 6-7) Reading for next lecture (Wed.): Chapter 6 (6-7 -- 6-10)Reading for next lecture (Wed.): Chapter 6 (6-7 -- 6-10) Homework for today's lecture: Chapter 6, question 13; problems 22, 30, 39, 41Homework for today's lecture: Chapter 6, question 13; problems 22, 30, 39, 41

2 Work and kinetic energy Work W= F || d =Fd cosθWork W= F || d =Fd cosθ Kinetic energy (KE) is defined to be KE=½mv 2Kinetic energy (KE) is defined to be KE=½mv 2 Units of both are in JoulesUnits of both are in Joules

3 The work-kinetic energy theorem Using calculus can prove W net =ΔKEUsing calculus can prove W net =ΔKE W net is the total work done by all the forces acting on the objectW net is the total work done by all the forces acting on the object ΔKE is final KE minus initial KEΔKE is final KE minus initial KE See p. 151 for proof in the special case of constant forcesSee p. 151 for proof in the special case of constant forces Proof comes from Newton's second law and the definitions of work and KEProof comes from Newton's second law and the definitions of work and KE ExampleExample

4 Your turn A 1 kg block goes 5 m down a frictionless incline that makes an angle of 37 o with the horizontal. If the block starts from rest, what is its final speed? Answer: As before, W = Fd cos(53 o ) = (1 kg)(9.8 m/s 2 )(5 m) cos(53 o )=29.5 J; so 29.5 J = KE 2 -KE 1, KE 1 =0, so 29.5 J=½mv 2 => v 2 = 2(29.5 J)/(1 kg) => v=7.7 m/s d=5 m 37 o

5 Potential energy (PE) Kinetic energy is associated with the object's motionKinetic energy is associated with the object's motion Potential energy is associated with conservative external forces acting on the bodyPotential energy is associated with conservative external forces acting on the body Change in potential energy is minus the work done by the force (ΔPE= - W)Change in potential energy is minus the work done by the force (ΔPE= - W) We'll discuss gravity and the elastic spring forceWe'll discuss gravity and the elastic spring force

6 Gravitational potential energy Raise ball from y 1 to y 2Raise ball from y 1 to y 2 Work done by gravity is W = -mg(y 2 - y 1 )Work done by gravity is W = -mg(y 2 - y 1 ) The ball has gained gravitational potential energyThe ball has gained gravitational potential energy ΔPE= - W; ΔPE= mg(y 2 - y 1 )ΔPE= - W; ΔPE= mg(y 2 - y 1 ) PE grav = mgyPE grav = mgy ExampleExample

7 Elastic spring force Hooke's Law: F S = -kxHooke's Law: F S = -kx k is the spring constantk is the spring constant x is the magnitude of the displacement from the spring's rest positionx is the magnitude of the displacement from the spring's rest position direction of the force is opposite the direction of the displacement xdirection of the force is opposite the direction of the displacement x

8 Elastic potential energy Force pulling the spring does work W P =½kx 2Force pulling the spring does work W P =½kx 2 W P = -W S = ΔPE=½kx 2W P = -W S = ΔPE=½kx 2 PE elastic =½kx 2PE elastic =½kx 2

9 Example A spring has constant k=500 N/m. a) It is stretched from rest to x=0.02 m. Find the PE, the work done to pull the spring and the work done by the spring. b) Continue stretching to x=0.04 m. Find the additional work done to pull the spring and the additional work done by the spring.

10 Example (cont.) A spring has constant k=500 N/m. c) Your turn: If the spring is compressed from x 1 =0 to x 2 =-0.02 m, find the PE, the work done to push the spring and the work done by the spring. Answer: W= ΔPE = PE 2 - PE 1 = ½kx 2 2 - ½kx 1 2 = ½(500 N/m)(-0.02 m) 2 =0.10 J is work to push spring; work done by spring is -ΔPE =-0.10J.

11 Conservative forces We cannot define PE for all forcesWe cannot define PE for all forces To define a PE, work must depend on initial and final positions and not the path in betweenTo define a PE, work must depend on initial and final positions and not the path in between Such a force is called "conservative"Such a force is called "conservative" Gravity and the elastic spring force are conservative forcesGravity and the elastic spring force are conservative forces Other conservative forces: electric force, Newton's universal gravityOther conservative forces: electric force, Newton's universal gravity

12 Non-conservative forces The path taken matters to the work doneThe path taken matters to the work done Cannot define a potential energy functionCannot define a potential energy function Mechanical energy is lost as heatMechanical energy is lost as heat Examples: friction, air resistanceExamples: friction, air resistance Compare work done by gravity, friction for two different paths from height yCompare work done by gravity, friction for two different paths from height y

13 Conservation of mechanical energy Recall this example --Recall this example -- A 1 kg block goes 5 m down a frictionless incline that makes an angle of 37 o with the horizontal. If the block starts from rest, what is its final speed?A 1 kg block goes 5 m down a frictionless incline that makes an angle of 37 o with the horizontal. If the block starts from rest, what is its final speed? Previous answer: W=mgy=29.5 J; so 29.5 J = KE 2 -KE 1, KE 1 =0, so 29.5 J = ½mv 2 => v 2 = 2(29.5 J)/(1 kg) => v=7.7 m/sPrevious answer: W=mgy=29.5 J; so 29.5 J = KE 2 -KE 1, KE 1 =0, so 29.5 J = ½mv 2 => v 2 = 2(29.5 J)/(1 kg) => v=7.7 m/s We want to rework this in the language of conservation of energyWe want to rework this in the language of conservation of energy

14 Conservation of energy approach Initially energy is all gravitational PEInitially energy is all gravitational PE As object falls PE is converted to KEAs object falls PE is converted to KE At the bottom, all the energy is kineticAt the bottom, all the energy is kinetic W = ΔKE = - ΔPE => (KE 2 - KE 1 ) = - (PE 2 - PE 1 ) => (KE 1 + PE 1 ) = (KE 2 + PE 2 )W = ΔKE = - ΔPE => (KE 2 - KE 1 ) = - (PE 2 - PE 1 ) => (KE 1 + PE 1 ) = (KE 2 + PE 2 ) E = KE + PE has the same value throughout the motionE = KE + PE has the same value throughout the motion E is "total mechanical energy"E is "total mechanical energy"

15 Revisiting the 1 kg block problem v 1 = 0, so KE 1 = 0v 1 = 0, so KE 1 = 0 y 1 = 3 m, so PE 1 = mgy 1 = (1 kg)(9.8 m/s 2 )(3 m) = 29.5 Jy 1 = 3 m, so PE 1 = mgy 1 = (1 kg)(9.8 m/s 2 )(3 m) = 29.5 J E = KE 1 + PE 1 = 29.5 JE = KE 1 + PE 1 = 29.5 J y 2 = 0, so PE 2 = 0y 2 = 0, so PE 2 = 0 E = KE 2 + PE 2 = 29.5 J => KE 2 = 29.5 J =½mv 2 => v 2 = 2(29.5 J)/(1 kg) => v=7.7 m/sE = KE 2 + PE 2 = 29.5 J => KE 2 = 29.5 J =½mv 2 => v 2 = 2(29.5 J)/(1 kg) => v=7.7 m/s A 1 kg block goes 5 m down a frictionless incline that makes an angle of 37 o with the horizontal. If the block starts from rest, what is its final speed? 37 o d=5 m 1 2

16 Pendulum energy analysis If no energy lost to friction, pendulum returns to same heightIf no energy lost to friction, pendulum returns to same height Energy all potential at top of swingEnergy all potential at top of swing Energy all kinetic at bottom of swingEnergy all kinetic at bottom of swing

17 Roller coaster example Consider the roller coaster in the figure. What speed is needed at point A to reach point B? Answer: E = KE A + PE A = KE B + PE B => ½mv 2 + mg(15 m) = mg(25 m) => ½mv 2 = mg(10 m) => v 2 = 2g(10 m) => v= 14.0 m/s Your turn: Find the speed at point C. Answer: E = KE B + PE B = KE C + PE C => mg(25 m) = ½mv 2 + mg(0 m) => ½mv 2 = mg(25 m) => v 2 = 2g(25 m) => v= 22.1 m/s

18 Conservation of energy If only conservative forces do work, then the total mechanical energy of the system remains constant.If only conservative forces do work, then the total mechanical energy of the system remains constant.


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