Consider an archery bow as behaving like a spring with a spring constant of 493 N/m. If an archer pulls the bowstring back for a distance of 42 cm, what is the elastic potential energy of the "loaded" bow?

If the mass of the arrow is 39 grams, what is its speed as it leaves the bow?

PE=1/2 kx^2 x=.42m

1/2 mv^2=1/2 kx^2
v=x sqrt(k/m)

To find the elastic potential energy of the loaded bow, we can use the formula:

Elastic potential energy = 0.5 * k * x^2

Where:
- k is the spring constant (493 N/m)
- x is the displacement of the bowstring (42 cm or 0.42 m)

Substituting these values into the formula:

Elastic potential energy = 0.5 * 493 N/m * (0.42 m)^2
Elastic potential energy = 52.26768 J (rounded to four decimal places)

Therefore, the elastic potential energy of the loaded bow is approximately 52.2677 Joules.

Now, let's calculate the speed of the arrow as it leaves the bow using the principle of conservation of mechanical energy. In this case, the elastic potential energy is converted into the kinetic energy of the arrow.

The kinetic energy formula is given by:

Kinetic energy = 0.5 * m * v^2

Where:
- m is the mass of the arrow (39 grams or 0.039 kg)
- v is the speed of the arrow

We can equate the elastic potential energy to the kinetic energy to find the speed:

Elastic potential energy = Kinetic energy

0.5 * k * x^2 = 0.5 * m * v^2

Simplifying and solving for v:

v^2 = (k * x^2) / m

v = sqrt((k * x^2) / m)

Substituting the given values:

v = sqrt((493 N/m * (0.42 m)^2) / (0.039 kg))
v = sqrt(56.1359 m^2/s^2)
v = 7.4963 m/s (rounded to four decimal places)

Therefore, the speed of the arrow as it leaves the bow is approximately 7.4963 m/s.