Work Energy Power Chapter 6 CJ Chapter 10Glencoe

Work, Energy, & Power Chapter 6 (C&J) Chapter 10(Glencoe)

Schedule for week of 2/3 -2/6 l Tuesday – lecture l l l Wednesday – Midterm Part I l l l Finish lecture In class worksheet, questions Thursday – Midterm Part II l l l Webassign due Sunday 11: 59 PM 2/8 In class worksheet , questions Finish lecture In class worksheet , questions Friday – class l In class worksheet , questions

Energy l What is energy? l l The capacity of a physical system to do work. What are some forms of energy? l l Kinetic Energy Potential Energy l l l l Gravitational Potential Energy (gravity) Elastic Potential Energy (springs, rubber bands) Chemical Energy (chemical bonds) Rest Mass Energy = Nuclear (E = mc 2) Electric Potential Energy (ΔU = kq 1 q 2/r) Thermal Energy (heat = KE of molecules) Sound (waves) Light (waves/photons)

Work l What is work? l Work is the application of a force to an object that causes it to move some displacement (d). W = Fd l Note: Work is a scalar quantity, i. e. it has magnitude, but no direction. F d

Kinetic Energy l Kinetic Energy is known as the energy of motion. n KE = ½ mv 2 If you double the mass, what happens to the kinetic energy? It doubles. l If you double the velocity, what happens to the kinetic energy? It quadruples. l

Kinetic Energy & Work l Newton’s 2 nd Law of Motion (Fnet = ma) n n Substituting n n vf 2 – vi 2 = 2 ad for a: v f 2 – v i 2 = Multiplying both sides of the equation by ½ m n ½ mvf 2 – ½ mvi 2 = Fnetd

Kinetic Energy & Work l The left side of the mathematical relationship is equal to the change in Kinetic Energy of the system. l KE = ½ mvf 2 – ½ mvi 2 l The right side of the mathematical relationship is equal to the amount of Work done by the environment on the system. l W = Fnetd

Work – Energy Theorem l The Work-Energy Theorem states that the work done on an object is equal to its change in kinetic energy. l l l ΔKE = W Note: this condition is true only when there is no friction. Units: l Joule (J) l l l 1 Joule is equal to the amount of work done by a 1 Newton force over a displacement of 1 meter. 1 Nm 1 kg • m 2/s 2

Calculating Work l What if the force is not completely in the same direction as the displacement of the object? F θ

Calculating Work When all the force is not in the same direction as the displacement of the object, we can use simple trig to determine the magnitude of the force in the direction of interest. l Hence: W = Fdcosθ l F θ Fx = Fcosθ Fy = Fsinθ

Example 1: l Little Johnny pulls his loaded wagon 30 meters across a level playground in 1 minute while applying a constant force of 75 Newtons. How much work has he done? The angle between the handle of the wagon and the direction of motion is 40°. F θ d

Example 1: l Formula: W = Fdcosθ l Known: Displacement: 30 m l Force 75 N l θ = 40° l Time = 1 minute l l Solve: l W = (75 N)(30 m)cos 40° = 1, 724 J

Example 2: The moon revolves around the Earth approximately once every 29. 5 days. How much work is done by the gravitational force? Gmmm. E l F= r 2 l (6. 67 x 10 -11 Nm 2/kg 2)(7. 35 x 1022 kg)(5. 98 x 1024 kg) l F= (3. 84 x 108 m)2 F = 1. 99 x 1020 N l In one lunar month, the moon will travel 2πr. E-m l l d = 2π(3. 84 x 108 m) = 2. 41 x 109 m

Example 2: W = Fdcosθ …… HOWEVER!! l Since: l l θ is 90°, Fcosθ = 0 While distance is large, displacement is 0, and Fd = 0 Hence: l W=0 d F

Work and Friction: Example 3 l 1. 2. 3. The crate below is pushed at a constant speed across the floor through a displacement of 10 m with a 50 N force. How much work is done by the worker? How much work is done by friction? What is the total work done? Ff F d = 10 m

Example 3 (cont. ): Wworker = Fd = (50 N)(10 m) = 500 J Wfriction = -Fd = (-50 N)(10 m) = -500 J If we add these two results together, we arrive at 0 J of work done on the system by all the external forces acting on it. l Alternatively, since the speed is constant, we know that there is no net force on the system. 1. 2. 3. l Since Fnet = 0, W = Fd = 0 Similarly, since the speed does not change: l l l Using the work-energy theorem we find that: W = ΔKE = ½ mvf 2 – ½ mvi 2 = 0.

Gravitational Potential Energy l If kinetic energy is the energy of motion, what is gravitational potential energy? l Stored energy with the “potential” to do work as a result of the Earth’s gravitational attraction and the objects position. l For example: l A ball sitting on a table has gravitational potential energy due to its position. When it rolls off the edge, it falls such that its weight provides a force over a vertical displacement. Hence, work is done by gravity.

Gravitational Potential Energy PE = mgΔh h Work By substituting Fg for mg, we obtain: PE = FgΔh Note: For objects close to the surface of the Earth: 1. g is constant. 2. Air resistance can be ignored.

Example 4: A 60 kg skier is at the top of a slope. By the time the skier gets to the lift at the bottom of the slope, she has traveled 100 m in the vertical direction. l 1. 2. If the gravitational potential energy at the bottom of the hill is zero, what is her gravitational potential energy at the top of the hill? If the gravitational potential energy at the top of the hill is set to zero, what is her gravitational potential energy at the bottom of the hill?

Case 1 PE = mgΔh m = 60 kg g = 9. 81 m/s 2 h = 100 m A PE = (60 kg)(9. 81 m/s 2)(100 m) PE = 59000 J PE = 59 k. J h = 100 m B

Case 2 PE = mgΔh m = 60 kg g = 9. 81 m/s 2 h = -100 m PE = (60 kg)(9. 81 m/s 2)(-100 m) PE = -59000 J PE = -59 k. J B h = 100 m A

Power l What l is it? Power is measure of the amount of work done per unit of time. P = W/t l What are the units? Joule/second l Watts l

Example 5: l Recalling Johnny in Ex. 1 pulling the wagon across the school yard. He expended 1, 724 Joules of energy over a period of one minute. How much power did he expend? P = W/t l P = 1724 J/60 s l P = 28. 7 W l

Alternate representations for Power l As l previously discussed: Power = Work / Time l Alternatively: l P = F d/t l l Since d/t = velocity P = F v l In this case here, we are talking about an average force and an average velocity.

Example 4: l. A corvette has an aerodynamic drag coefficient of 0. 33, which translates to about 520 N (117 lbs) of air resistance at 26. 8 m/s (60 mph). In addition to this frictional force, the friction due to the tires is about 213. 5 N (48 lbs). l Determine the power output of the vehicle at this speed.

Example 4 (cont. ) l The total force of friction that has to be overcome is a sum of all the external frictional forces acting on the vehicle. l l l Ff = Fair drag + Ftire resistance Ff = 520 N + 213. 5 N = 733. 5 N P = F v l l l P = (733. 5 N)(26. 8 m/s) = 19, 657. 8 W P = 26. 4 hp If an engine has an output of 350 hp, what is the extra 323. 6 horsepower needed for? l l Acceleration Plus, at higher speeds the resistive forces due to air and tire friction increase.

Key Ideas Energy of motion is Kinetic Energy = ½ mv 2. l Work = The amount of energy required to move an object from one location to another. l The Work-Energy Theorem states that the change in kinetic energy of a system is equal to the amount of work done by the environment on that system. l Power is a measure of the amount of work done per unit of time. l

Review l Energy l l l KE = ½ mv 2 PE = FgΔh = mgΔh Work results in a change in energy l W = Fdcosθ KE = ½ mvf 2 – ½ mvi 2 = Fnetd l Power = P = W/t = F v l

Review l Kinetic Energy l KE = ½ mv 2 l. A change in energy requires work. Work = F*d l DKE = ½ m(vf 2 – vi 2 ) = ½ m (2 ad) = (ma)d = F*d = Work l l PE = F*d = (mg)h = F*d