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# Purpose

• To determine the efficiency of a homemade recurve bow to transfer its initial energy into its final energy
• To better understand efficiency with energy transfer

# Work and Energy

Work is when a force acts upon an object to cause a displacement of the object. On the other hand, there are 2 basic forms of energy (potential energy and kinetic energy) associated with shooting an arrow. Potential energy is the stored energy of position possessed by an object. Additionally, potential energy can be mathematically modeled through the equation: mass(kg)*gravity(m/s)*height(m). Kinetic energy is the energy of motion. Similar to potential energy, kinetic energy can be mathematically modeled through the equation: 1/2 mass(kg)* velocity^2(m/s).

# Energy Model

The energy transfer model for shooting an arrow can be illustrated by the equation:

efficiency(work + potential energy) = (kinetic energy + potential energy)

e(Fd + mgy) = 1/2mv^2 + mgy

# Work

To begin, I set up this experiment to observe and record the relationship between drawback force and distanced pulled. I conducted this experiment by using a force sensor to read and record the drawback force in increments of 5 centimeters.

Logically, the graph of drawback force as a function of distance appeared to be exponential.

# Potential Energy

To determine the actual drawback force of my shot, I used video analysis of the starting position of the string, indicated by the solid vertical yellow line, and the final position of the string, indicated by the blue dot, to find my drawback force of .3491 meters or 34.91 centimeters.

To determine the work done on the string by pulling it back, it is necessary to find the area under the exponential curve. This graph shows drawback force in Newtons versus distance in cm, thus when converted, the work done on the system is 5.196 Joules when the string is displaced .3491 meters or 34.91 centimeters.

To determine the initial potential energy of the arrow, it was essential to figure out how high the arrow was off the ground. Using the potential energy equation, m*g*h, I plugged in the mass of the arrow, the gravitational constant, and 1.527 meters to calculate the initial potential energy of the arrow.

On the other hand, to determine the final potential energy of the arrow, I once again video analyzed the projectile motion of the arrow. As you can see in block 26 of the data table, the y position is .1644 meters lower than the starting position. Thus, I subtracted the initial y position by .1644 meters to determine the arrow's height off the ground. Then, using this height, the mass of the arrow, and the gravitational constant, I was able to calculate the arrow's final potential energy.

*Check "Calculations and Extra Data" section at the bottom of the presentation*

# Kinetic Energy

To find out the final kinetic energy of the arrow, I had to use the x and y velocities of the arrow of block 26 of the data table and a little bit of trigonometry to find the velocity at which the arrow strikes the target. Once I calculated the final velocity of the arrow, I could plug in the known arrow mass and the final velocity into the kinetic energy formula, 1/2 mass(kg)* velocity^2(m/s), to determine the arrow's final kinetic energy.

*Check "Calculations and Extra Data" section at the bottom of the presentation*

# Efficiency

Through all these calculations, I determined that the efficiency of my homemade recurve bow to transfer its initial energy to its final energy is 52.74%. In simpler terms, the energy that the arrow had prior to launch only converted/transferred 52.74% of its energy into its final kinetic and potential energies. This is a reasonable amount of efficiency because there are several factors that contribute to inefficient energy transfer such as:

• The arrow rubbing on top of my hand during launch
• The fetching of the arrows caught/rubbed against my hand and bow
• Air resistance
• Drag
• Sound
• Heat
• Low budget/quality materials (pvc bow, cheap arrows)

*Check "Calculations and Extra Data" section at the bottom of the presentation*

# Application of Efficiency

How can velocity be calculated using the efficiency of a bow to transfer its energy? Using the same scenario, how will velocity be impacted if the work done on the system is tripled and the original arrow is replaced with a high tech carbon fiber arrow that is 1/3 the mass of the original arrow?

Answer = 46.377 m/s which is about 3 times faster than the original velocity

*Check "Calculations and Extra Data" section at the bottom of the presentation*

# Conclusion

• Area under curve of drawback force as a function of distance = work done on system
• e(Fd + mgy) = ½ mv2 + mgy
• Initial energy ≠ Final energy if factors are not accounted for

# Calculations and Extra Data

For detailed calculations and extra data please view the following Google doc: