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PHSX 114, Monday, September 29, 2003 • Reading for today: Chapter 6 (6-3 -- 6-7) • Reading for next lecture (Wed.): Chapter 6 (6-7 -- 6-10) • Homework for today's lecture: Chapter 6, question 13; problems 22, 30, 39, 41 Work and kinetic energy • Work W= F||d =Fd cosθ • Kinetic energy (KE) is defined to be KE=½mv2 • Units of both are in Joules The work-kinetic energy theorem • Using calculus can prove Wnet=ΔKE • Wnet is the total work done by all the forces acting on the object • ΔKE is final KE minus initial KE • See p. 151 for proof in the special case of constant forces • Proof comes from Newton's second law and the definitions of work and KE • Example d=5 m Your turn 37o A 1 kg block goes 5 m down a frictionless incline that makes an angle of 37o with the horizontal. If the block starts from rest, what is its final speed? Answer: As before, W = Fd cos(53o) = (1 kg)(9.8 m/s2)(5 m) cos(53o)=29.5 J; so 29.5 J = KE2-KE1, KE1=0, so 29.5 J=½mv2 => v2= 2(29.5 J)/(1 kg) => v=7.7 m/s Potential energy (PE) • Kinetic energy is associated with the object's motion • Potential 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) • We'll discuss gravity and the elastic spring force Gravitational potential energy • Raise ball from y1 to y2 • Work done by gravity is W = -mg(y2 - y1) • The ball has gained gravitational potential energy • ΔPE= - W; ΔPE= mg(y2 - y1) • PEgrav= mgy • Example Elastic spring force • Hooke's Law: FS = -kx • k is the spring constant • x is the magnitude of the displacement from the spring's rest position • direction of the force is opposite the direction of the displacement x Elastic potential energy • Force pulling the spring does work WP=½kx2 • WP= -WS= ΔPE=½kx2 • PEelastic=½kx2 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. Example (cont.) A spring has constant k=500 N/m. c) Your turn: If the spring is compressed from x1=0 to x2=-0.02 m, find the PE, the work done to push the spring and the work done by the spring. Answer: W= ΔPE = PE2- PE1= ½kx22 - ½kx12= ½(500 N/m)(-0.02 m)2=0.10 J is work to push spring; work done by spring is -ΔPE =-0.10J. Conservative forces • We cannot define PE for all forces • To define a PE, work must depend on initial and final positions and not the path in between • Such a force is called "conservative" • Gravity and the elastic spring force are conservative forces • Other conservative forces: electric force, Newton's universal gravity Non-conservative forces • • • • • The path taken matters to the work done Cannot define a potential energy function Mechanical energy is lost as heat Examples: friction, air resistance Compare work done by gravity, friction for two different paths from height y Conservation of mechanical energy • Recall this example -• A 1 kg block goes 5 m down a frictionless incline that makes an angle of 37o 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 = KE2-KE1, KE1=0, so 29.5 J = ½mv2 => v2= 2(29.5 J)/(1 kg) => v=7.7 m/s • We want to rework this in the language of conservation of energy Conservation of energy approach • • • • Initially energy is all gravitational PE As object falls PE is converted to KE At the bottom, all the energy is kinetic W = ΔKE = - ΔPE => (KE2 - KE1) = - (PE2 - PE1) => (KE1 + PE1) = (KE2 + PE2) • E = KE + PE has the same value throughout the motion • E is "total mechanical energy" Revisiting the 1 kg block problem A 1 kg block goes 5 m down a frictionless incline that makes an angle of 37o with the horizontal. If the block starts from rest, what is its final speed? • • • • • 1 d=5 m 37o v1= 0, so KE1 = 0 y1= 3 m, so PE1 = mgy1= (1 kg)(9.8 m/s2)(3 m) = 29.5 J E = KE1 + PE1 = 29.5 J y2= 0, so PE2 = 0 E = KE2 + PE2 = 29.5 J => KE2 = 29.5 J =½mv2 => v2= 2(29.5 J)/(1 kg) => v=7.7 m/s 2 Pendulum energy analysis • If no energy lost to friction, pendulum returns to same height • Energy all potential at top of swing • Energy all kinetic at bottom of swing Roller coaster example Consider the roller coaster in the figure. What speed is needed at point A to reach point B? Answer: E = KEA + PEA = KEB + PEB => ½mv2 + mg(15 m) = mg(25 m) => ½mv2 = mg(10 m) => v2 = 2g(10 m) => v= 14.0 m/s Your turn: Find the speed at point C. Answer: E = KEB + PEB = KEC + PEC => mg(25 m) = ½mv2 + mg(0 m) => ½mv2 = mg(25 m) => v2 = 2g(25 m) => v= 22.1 m/s Conservation of energy • If only conservative forces do work, then the total mechanical energy of the system remains constant.