Energy
![Picture](/uploads/1/5/1/5/15153444/8385607.jpg?367)
Energy is the ability or capacity to do work. There are two main forms of energy: the first form is Kinetic which is energy that is possessed by moving objects. The second form of energy is Potential energy which is a form of energy that an object possess because of its position in relation to forces in it's environment such as the force of gravity. We use energy in everything we do and we do not realize that in our everyday lives. We convert energy from one form to another so that we can power different machines and vehicles. In American football energy is used in the game of football from when ball is resting still in the players hand to when the ball is released from the players hands. The ball will always have gravitational potential energy when it is not in motion. When the ball is kicked of thrown that gravitational potential energy has now transformed to kinetic energy. Let take a closer look at the energy a ball has when in the hands of Michael Vick. We will also investigate the energy of the ball when Peyton manning puts the ball in motion by throwing it.
Foot ball in his hands
![Picture](/uploads/1/5/1/5/15153444/8207847.jpg?368)
Gravitational Potential Energy:
Michael is Vick is 1.83m so lets assume the height he holds the football in his hands is around 1.40m.
Given: Mass of ball = (0.400kg) g = (9.8m/s^2) height = (1.40m)
Required: Eg
Analysis: Eg = mgh
Substitute: Eg = (0.400kg)(9.8m/s^2)(1.40m)
Solution: Eg = 5.49 J
Therefore the gravitational potential energy of the ball
Michael is Vick is 1.83m so lets assume the height he holds the football in his hands is around 1.40m.
Given: Mass of ball = (0.400kg) g = (9.8m/s^2) height = (1.40m)
Required: Eg
Analysis: Eg = mgh
Substitute: Eg = (0.400kg)(9.8m/s^2)(1.40m)
Solution: Eg = 5.49 J
Therefore the gravitational potential energy of the ball
Foot ball in the air
![Picture](/uploads/1/5/1/5/15153444/7667590.jpg?301)
Kinetic Energy:
Peyton manning can throw a NFL football that is will travel at a constant velocity of 28m/s.
Given: Mass of he ball = (.400kg) velocity = (28 m/s^2)
Required: Ek
Analysis: Ek = m(v)^2 / 2
Substitute: Ek = (.400kg) (28m/s^2)^2 / 2
Solution: Ek = 156.8 J
Therefore the Kinetic energy of the ball in motion is 156.8 J
Peyton manning can throw a NFL football that is will travel at a constant velocity of 28m/s.
Given: Mass of he ball = (.400kg) velocity = (28 m/s^2)
Required: Ek
Analysis: Ek = m(v)^2 / 2
Substitute: Ek = (.400kg) (28m/s^2)^2 / 2
Solution: Ek = 156.8 J
Therefore the Kinetic energy of the ball in motion is 156.8 J
Law of conservation of energy
![Picture](/uploads/1/5/1/5/15153444/7302374.png?408)
This law of energy states the energy can not be created or destroyed. All the energy in a particular system will remain constant throughout the whole trip. This law is represented with the equation of Et = Ek + Ep. Using this equation we can find the total energy that the ball has for the journey of the flight.
So lets say a quarter back hold the football at a height of 1.40m and the 0.400kg NFL football is thrown with a velocity of 35 m/s^2.
We need use the law of conservation of energy to find energy total.
Given: m = (0.400kg) v = (35m/s^2) g = (9.8m/s^s)
Required: Et
Analysis: Et = Ek + Eg
Substitute: Et = (0.400kg) (35m/s^2)^2 / 2 + (0.400kg) (9.8m/s^2) (1.40m)
Solution: Et = 245 + 5.48
= 254 J
Therefore energy total is 254 J
So lets say a quarter back hold the football at a height of 1.40m and the 0.400kg NFL football is thrown with a velocity of 35 m/s^2.
We need use the law of conservation of energy to find energy total.
Given: m = (0.400kg) v = (35m/s^2) g = (9.8m/s^s)
Required: Et
Analysis: Et = Ek + Eg
Substitute: Et = (0.400kg) (35m/s^2)^2 / 2 + (0.400kg) (9.8m/s^2) (1.40m)
Solution: Et = 245 + 5.48
= 254 J
Therefore energy total is 254 J
Work
![Picture](/uploads/1/5/1/5/15153444/349594.jpg?306)
Work is the transfer of energy such as applying a force on an object that displaces the object in the same direction as the force applied. Work is a scalar quantity which means that there is no direction associated with work. There are two type of work: Positive and Negative. Positive work is when the force and displacement are in the same direction. Negative work is when the force and displacement are in the opposite direction. In this case with the football thrown positive work is being done. Work is represented by the equation W = Fd as long as the force and displacement are in the same direction. We can use this equation to out how much work Peyton Manning does on the football. Peyton manning 1.96 m tall, weights 104.3 kg, and applies a force of 600N to throw the ball 25m. With this imformation we can now determine the amount of work that Peyton has done.
Given: Force applied = 600N [Forward] Distance thrown [25m]
Required: W
Analysis: W = Fd
Substitute: W = (600N)(25m)
Solution: W = 1.5 x10^4 J
Therefore Peyton does 15000 J of mechanical work to move the object 25m.
Given: Force applied = 600N [Forward] Distance thrown [25m]
Required: W
Analysis: W = Fd
Substitute: W = (600N)(25m)
Solution: W = 1.5 x10^4 J
Therefore Peyton does 15000 J of mechanical work to move the object 25m.
Power
![Picture](/uploads/1/5/1/5/15153444/1353967202.jpg)
Power is the rate or measurement of transforming energy or doing work. You can calculate power with the equation P = Wnet/t. Power is measured with the standard metric unit "Watts" represented by a "W". In the game of American football players produce tremendous power when tackling opponents. We can find the amount of power that is produced when a player by using the mass and the acceleration of the player to find the change in kinetic energy. We then divide the change in energy by the total time the player is running for. For an example Michael Vick runs the 40 yard dash in 4.36 seconds if we change yards to meter that would be approximately 37m. His average velocity is 8.5m/s^2.
First we have to find the kinetic energy he produces. Ek1 is 0 because you start from rest. So you have to find Ek2
Given: Mass of Michael Vick = 98kg V = 8.5m/s^2
Required: Ek
Analysis: Ek = mv^2/ 2
Substitute: Ek = (98kg) (8.5)^2 / 2
Solution: Ek = 3540.25 J
Now we use delta E and plug it into our power equation. We are assuming that the player is running for 3 seconds before he makes the tackle.
Given: Delta E = 3540.25 Delta t = 3.0
Required: Power
Analysis: P = E/t
Substitute: P = 3540.25/3.0
Solution: 1180.08
Therefore Vick can produce 1180.08 watts of power when he tackles a player.
First we have to find the kinetic energy he produces. Ek1 is 0 because you start from rest. So you have to find Ek2
Given: Mass of Michael Vick = 98kg V = 8.5m/s^2
Required: Ek
Analysis: Ek = mv^2/ 2
Substitute: Ek = (98kg) (8.5)^2 / 2
Solution: Ek = 3540.25 J
Now we use delta E and plug it into our power equation. We are assuming that the player is running for 3 seconds before he makes the tackle.
Given: Delta E = 3540.25 Delta t = 3.0
Required: Power
Analysis: P = E/t
Substitute: P = 3540.25/3.0
Solution: 1180.08
Therefore Vick can produce 1180.08 watts of power when he tackles a player.
COnduction
![Picture](/uploads/1/5/1/5/15153444/1353957511.png)
Conduction is one of the few types of heat transfer. Conduction is the transfer of thermal energy that occurs when warmed objects are in contact with colder objects. The diagram on the right shows an example of conduction as the stove is touching the pan. The particles in the in the pan are cold but as the stove touches it the heat from the stove causes the particles to speed up. Although the stove seems to stay hot the stove particles are slowing down but at a slower rate than the pan is heating up. This can be applied in football because as the player hold the football in his hands the heat from his hands are being transferred to the ball. We can find the final temperature of the hand and ball using Qreleased + Qabsorbed = 0.
For an example we have a 65kg quarterback who is 37 degrees Celsius before touching 0.400kg football with an initial temperature of 20 degrees Celsius.
Qreleased:
Given: m = 5kg c = 4.18x10^3 T=(Tf - 37degrees Celsius)
Qr = (5kg) (4.18x10^3) (TF-37)
= 20900TF-773300 degrees Celsius
Qabsorbed:
Given: m = .400kg c = 1.5x10^3 T= (Tf - 20 degrees Celsius)
Qr = (0.400kg) (1.5x10^3) (TF-20 degrees Celsius)
= 600TF - 12000 degrees Celsius
Qreleased + Qabsorbed = 0
20900TF - 773300 + 600TF - 12000 degrees Celsius = 0
21500Tf = 785300
Tf = 36
Therefore the final temperature of the hand and ball is 36 degrees Celsius.
For an example we have a 65kg quarterback who is 37 degrees Celsius before touching 0.400kg football with an initial temperature of 20 degrees Celsius.
Qreleased:
Given: m = 5kg c = 4.18x10^3 T=(Tf - 37degrees Celsius)
Qr = (5kg) (4.18x10^3) (TF-37)
= 20900TF-773300 degrees Celsius
Qabsorbed:
Given: m = .400kg c = 1.5x10^3 T= (Tf - 20 degrees Celsius)
Qr = (0.400kg) (1.5x10^3) (TF-20 degrees Celsius)
= 600TF - 12000 degrees Celsius
Qreleased + Qabsorbed = 0
20900TF - 773300 + 600TF - 12000 degrees Celsius = 0
21500Tf = 785300
Tf = 36
Therefore the final temperature of the hand and ball is 36 degrees Celsius.
Fooood
![Picture](/uploads/1/5/1/5/15153444/2398620.jpg?1)
Football is a very high intensity game which required lots of energy from the players. The players are always in motion and moving whether it is running to catch the ball, tackling a player, or kicking a football. This means the players must have the proper fuel to play the game or else they will not perform the way they want to. To get this proper fuel players need right kind of food because in food are calories and these calories is a source of energy that our body uses to function. These foods must have good calories and what I mean by that is no trans fat, low or no saturated fats, protein and carbohydrates to provide enough energy to last throughout the game. For an example a good pre game meal can be a couple eggs, fresh fruit, toast, and some whole wheat pasta, and a light salad can be about 700Cal, totalling 2930.06 Joules of energy (convertunits.com) That is enough energy to last you through the game. Don't forget to have a proper meal after the game to refuel up again!
Calculations
1 Cal = 4.19 Joules
700 Cal x 4.19J = 2930.06 J
Calculations
1 Cal = 4.19 Joules
700 Cal x 4.19J = 2930.06 J