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Cultural Connections to Potential and Kinetic Energy

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    50731
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    From 1337 until 1453, the 100 years war was fought between the English and the French. They fought to determine the who the new king of France would be. Since the Normans (from Normandy, France) had conquered England much earlier on, the English believed they had the right to the French throne. The French believed that they should hold the throne as they were French. In 1337 the fighting began with naval battles in the English channel. The British won these battles, ensuring that the war would continue on French soil, rather than English. Small skirmishes continued until the Battle of Crécy in 1346. In this battle King Edward III of England found himself out-maneuvered by the French King Philip. In the end, King Edward III relied upon the supremacy of his archers in battle.

    Figure \(\PageIndex{1}\) English Longbow Archers at the Battle of Crecy.

    The English archers were equipped with the English Longbow. This Longbow took many years to master, and many soldiers began training when they were young. At the time, the law in England required all men to be capable of shooting a Longbow. The bow was not easy to shoot. It was over 6 feet long, and had a pulling force of 180-200 pounds. That is an incredible pulling force. In comparison, the majority of today's long bows have a range of 40-80 pounds of pulling force. These men had to be extremely strong. The pulling force also shows the deadliness of the bow. At 200 pounds of pulling force, the arrow could easily penetrate French armor. The French did not rely on the long bow, but rather the cross bow. The cross bow was significantly worse because it took longer to load, was not as accurate, and did not have as much force behind it. However, any common man could shoot a crossbow, but only a highly trained soldier could shoot a longbow. At the Battle of Crécy shown in Figure 1, the English used their advantage with the Longbow to decimate the French troops and win the battle. It was a decisive victory in the beginning of the 100 years war that was made possible by the effectiveness of the English Longbow.

    The supremacy of the English Longbow over the cross bow is a great example of the effects of potential and kinetic energy. When the Longbow is pulled back, with the arrow cocked, this gives the bow potential energy. The string is being moved from its original position and therefore has the desire to return to it. It is the same idea as when someone holds a ball in an outstretched arm. The ball is held above the ground, yet gravity dictates that it should fall. The energy stored in the ball by the act of lifting the ball up effectively increases the ball's potential energy. The ball has potential energy just like the Longbow has when it is stretched. The other aspect of energy that the bow portrays is kinetic energy. Kinetic energy is the energy of motion. When the string is released, the potential energy is converted to kinetic energy in movement of the arrow. The string pushes the arrow out and is the driving force behind the motion. The harder an archer works to pull the bow, the greater the potential energy, and thus the greater the kinetic energy of the arrow.

    Potential Energy

    Potential Energy is energy that is stored by rising in height, or by other means. It frequently comes from separating things that attract, like holding a ball above the Earth that attracts it, or the bow string being pulled back when the wood part of the bow opposes it. Potential Energy is abbreviated EP and gravitational potential energy is calculated as follows:

    \[E_{P} = mgh\]

    Where

    • m = mass of the object in kg
    • g = gravitational constant, 9.8 m s2
    • h = height in m, or draw length

    Notice that EP has the same units, kg m2 s–2 or Joule, as kinetic energy. A Joule is also equal to N • m.

    A Longbow is more like a spring because as it is drawn back the bow bends, just like a spring, which follows Hooke's Law:

    \[ E_{P} = \dfrac{1}{2}kx^{2}\]

    Where

    • k = the force constant of the bow (or spring). (in N/m)
    • x = the distance the bow (or spring) is pulled. (in m)
    Figure \(\PageIndex{2}\). Drawn Bow with Pulling Distance Shown

    Example \(\PageIndex{1}\): Potential Energy of an Arrow

    How much potential energy would an arrow have fully pulled back in an 800N (180lb) English Longbow if the arrow weighed 880g and was drawn back 762 cm?

    Solution

    Using the equation above, we find that k = 800N (pulling force of bow), and x = 0.762m (draw length)

    Putting them all together we get that

    \[ \text{E}_{P} = \dfrac{1}{2}kx^{2} = \dfrac{1}{2} (\dfrac{800}{0.762}) \text{(0.762)}^2 = \text{304.8 J}\)]

    of Potential Energy

    You may be wondering why the mass of the arrow was not included. It was not included because we are doing the potential energy of the bow. As stated above, the bow acts like a spring, and therefore the mass of the arrow does not matter. Only the draw length and drawing force of the bow matter

    Kinetic Energy

    Energy due to motion is called kinetic energy and is represented by Ek. For an object moving in a straight line, the kinetic energy is one-half the product of the mass and the square of the speed:

    \[ \large E_{k} = \dfrac{1}{2}mu^{2} \]

    Where

    m = mass of the object

    u = speed of object

    Figure \(\PageIndex{3}\). English Longbow

    If the arrow mentioned is released from the bow it then has kinetic energy. Assuming there is no air resistance or loses due to friction with the bow, etc. The energy should stay constant from the potential energy through to the kinetic energy. This is also called the law of conservation of energy, where energy is not destroyed or lost, simply transformed. Here though, two different bows and their respective speeds are being used so the energy's will not be equal, although they are close, but it is something to keep in mind.

    Example \(\PageIndex{2}\): Kinetic Energy of an Arrow

    Calculate the kinetic energy of an arrow launched from an English Longbow weighing .20kg and launched at a speed of 200 f/s?

    Solution

    m = .20kg and u = 60.96 m/s then can be applied to the equation above:

    \(\text{E}_k = \dfrac{1}{2} \text{(0.20)(60.96)}^2 = \text{371.6 J}\)

    From ChemPRIME: 3.6: Energy

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