Power and Energy

James Joule British physicist James Joule is best known for his work in electricity and thermodynamics Together with the physicist William Thomson (later Baron Kelvin), Joule found that the temperature of a gas falls when it expands without doing any work. This principle, which became known as the Joule-Thomson effect, underlies the operation of common refrigeration and air conditioning systems. The metric system unit of energy is the joule (J), after James Joule.

Energy Facts There are different types of energy Energy of all types is measured in Joules Law of Conservation of Energy – Energy can be neither created nor destroyed, merely changed from one form to another

Types of Energy (Unit Overview) Mechanical Potential Energy Energy of Position Gravitational Elastic Kinetic Energy Energy of Motion If it moves it has kinetic energy Heat Energy Heat is a form of Energy Transfer Other Forms of Stored Energy Chemical Fuels - usually release energy by combustion Food – energy released by digestion Electrical Generated from other forms of energy

Work The Physics definition of work requires a displacement, i.e. an object must be moved in order for work to be done! The Applied force which causes the displacement contributes to the work, i.e. in order to contribute to the work, the applied force must be parallel to the displacement.

Work: A Mathematical Definition Work = (Force)(Displacement)(cos Ɵ ) Units of Work = (Newton)(Meter) 1 NewtonMeter = 1 Joule A Joule is a unit of Energy and it takes energy to do work and work done on an object either causes it to move (kinetic energy) or is stored (potential energy)

Sample Problem What work is done sliding a 200 Newton box across the room if the frictional force is 160 Newtons and the room is 5 meters wide? W = F f Δd = (160 N)(5 m) 800 Joules

Kinetic Energy Kinetic Energy is energy of Motion Any moving object has kinetic energy Dependent on the mass of the object and its velocity. Mathematically expressed as: E k = ½ mv 2

Sample Problem What is the kinetic energy of a car with a mass of 2000 kg moving at 30 m/s? E k = ½ mv 2 = (½)(2000 kg)(30 m/s) 2 = 900,000 Joules

Energy of Position: Gravitational Potential Energy Occurs due to the force of gravity Is determined by the position of the object in the gravitational field Mathematically determined by: E p = mgh where m is mass, g is the acceleration due to gravity and h is the height above a determined baseline.

Sample Problem What is the potential energy of a 10 kg rock sitting on a cliff 30 meters high? The acceleration due to gravity is 9.8 m/s 2. E p = mgh = (10 kg)(9.8 m/s 2 )(30 m) 2940 Joules

Elastic Potential Energy Bungee cords, rubber bands, springs any object that has elasticity can store potential energy. Each of these objects has a rest or “zero potential” position When work is done to stretch or compress the object to a different position elastic potential energy is stored

Elastic Potential Energy Top picture is “rest position”; x = 0 This is a point where the elastic potential energy = 0 Bottom picture is “stretched position” Here elastic potential energy is stored in the spring E p = ½ kx 2 where k is the “spring constant” in N/m

Sample Problem What is the Elastic potential energy of a car spring that has been stretched 0.5 meters? The spring constant for the car spring is 90 N/m. E p = ½ kx 2 = (½)(90 N/m)(0.5 m) 2 =11.25 Joules

Where Does “K” Come From? K is measured in Newtons/meter. It is defined as the force required to displace a spring 1 meter. So: K = F/x Often K is determined by hanging a known weight from the spring and measuring how much it is stretched from its rest postion.

Sample Problem A spring is hung from a hook and a 10 Newton weight is hung from the spring. The spring stretches 0.25 meters. What is the spring constant? If this spring were compressed 0.5 meters, how much energy would be stored? If this spring were used to power a projectile launcher, which fires a 0.2 kg projectile, with what velocity would the projectile leave the launcher? Assume 0.5 m compression.

Solution K = F/x K =10 N/0.25 m = 40 N/m E p = ½ Kx 2 E p = ½ (40 N/m)(0.5 m) 2 = 5 Joules E p = E k = ½ mv 2 5 Joules = ½ (0.2 kg)(v 2 ) V = 7.05 m/s

Power Power = Work/time = Joules/Second Mathematically there are two formulas for Power: or since then

Sample Problem What power is developed by a 55 kg person who does 20 chin ups, h = 3 m, in 45 seconds? P= w/t = FΔd/t = mgh/t (20(55 kg)(9.8 m/s 2 )(3 m))/45 sec = Watts

Problem Types Work Work at an angle Kinetic Energy Gravitational Potential Elastic Potential Conservation Power