a full draw is the maximum distance that an archer can pull an arrow. Using the recurve bow shown below, a particular archer requires an average force of 130N to pull a full draw of 70.0cm. (when the arrow is relaxed in the bow, there is a distance of 30.0cm btween the arrow touching the back string and the middle of the arcc. When the arrow is stretched, the distance is now increased by 700.0cm to a total of 100.0 cm).

1. How much potential energy is stored in the bow?

2. If the arrow has a mass of 18.5 g(0.0186kg), what is the maximum speed that the arrow could leave the bow with(assuming all potential energy is transformed into the kinetic energy of the arrow)?

3. If the arrow only leaves the bow with a speed of 54m/s, how much energy was lost to heat?

To answer these questions, we need to understand the principles of potential energy, kinetic energy, and work. Here's how we can approach each question:

1. How much potential energy is stored in the bow?

The potential energy stored in the bow can be calculated using the formula: Potential Energy = (1/2) * k * x^2

Here, k represents the spring constant of the bow, and x represents the displacement of the bowstring from its equilibrium position. To find the potential energy, we need to determine k and x values.

The given information states that a force of 130N is required to pull a full draw of 70.0cm (0.70m). Since force (F) is related to the spring constant (k) and displacement (x) by the formula F = k * x, we can rearrange the equation to solve for k: k = F / x

Substituting the given values:
k = 130N / 0.70m = 185.71 N/m

The displacement of the bowstring is given as the increase from 30.0cm (0.30m) to 100.0cm (1.00m), so the actual x value is 0.70m.

Using the formula for potential energy, we can calculate:
Potential Energy = (1/2) * k * x^2 = (1/2) * 185.71 N/m * (0.70m)^2 = 45.50 Joules

Therefore, the potential energy stored in the bow is 45.50 Joules.

2. If the arrow has a mass of 0.0186kg, what is the maximum speed that the arrow could leave the bow with (assuming all potential energy is transformed into the kinetic energy of the arrow)?

The potential energy stored in the bow is converted into kinetic energy of the arrow when it is released. To find the maximum speed of the arrow, we can equate the potential energy to the kinetic energy and solve for the velocity.

The formula for kinetic energy is given by: Kinetic Energy = (1/2) * m * v^2

Setting the potential energy equal to the kinetic energy:
Potential Energy = Kinetic Energy
45.50 Joules = (1/2) * 0.0186kg * v^2

Solving for v:
v^2 = (2 * 45.50 Joules) / 0.0186kg
v^2 = 4897.85 m^2/s^2
v ≈ 69.99 m/s

Therefore, the maximum speed that the arrow could leave the bow with is approximately 69.99 m/s.

3. If the arrow only leaves the bow with a speed of 54m/s, how much energy was lost to heat?

To find the energy lost to heat, we need to calculate the difference between the initial potential energy of the bow and the final kinetic energy of the arrow.

Using the formula for kinetic energy: Kinetic Energy = (1/2) * m * v^2
Kinetic Energy = (1/2) * 0.0186kg * (54m/s)^2 = 32.15 Joules

The energy lost to heat is the difference in energy:
Energy lost to heat = Potential Energy - Kinetic Energy
Energy lost to heat = 45.50 Joules - 32.15 Joules = 13.35 Joules

Therefore, the energy lost to heat is 13.35 Joules.