An archer pulls the bowstring back for a distance of 0.400 m before releasing the arrow. The bow and string act like a spring whose spring constant is 490 N/m.

(a) What is the elastic potential energy of the drawn bow?
(b) The arrow has a mass of 0.0300 kg. How fast is it traveling when it leaves the bow?

a) Elastic P.E. = (1/2) k X^2, where

k = 490 N/m and X = 0.400 m

b) Set Elastic P.E = (1/2) M v^2 and solve for V.

kX^2 = mV^2
V = X sqrt (k/m)

Well, it seems like the archer is quite "drawn" to springy situations. Let's dive into the hilarious world of elastic potential energy and arrow velocity!

(a) To calculate the elastic potential energy, we can use the formula:

Elastic Potential Energy = (1/2) * k * x^2

where k is the spring constant and x is the displacement of the bowstring.

So, with a spring constant of 490 N/m and a displacement of 0.400 m, plugging these values in:

Elastic Potential Energy = (1/2) * 490 N/m * (0.400 m)^2

Wait for it... *drumroll*... the answer is 39.2 Joules of elastic potential energy!

(b) Now let's find out how fast that arrow is zooming away from the bow. To calculate its velocity, we can use the concept of conservation of mechanical energy. The initial elastic potential energy will be converted into kinetic energy as the arrow is released.

The kinetic energy of the arrow can be calculated using the formula:

Kinetic Energy = (1/2) * m * v^2

where m is the mass of the arrow and v is the velocity.

We already know the mass of the arrow is 0.0300 kg, so let's plug that in along with the elastic potential energy we calculated earlier:

39.2 J = (1/2) * 0.0300 kg * v^2

Solving this equation gives us... *suspense*... the velocity of the arrow is approximately 28.1 m/s!

So, there you have it! The drawn bow has an elastic potential energy of 39.2 Joules, and the arrow leaves the bow at a speed of around 28.1 m/s. That's one speedy arrow, ready to hit bullseyes and tickle funny bones!

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

Elastic Potential Energy = (1/2) * k * x^2

Where:
k is the spring constant
x is the displacement (distance the bowstring is pulled back)

Given:
k = 490 N/m
x = 0.400 m

(a) Calculate the elastic potential energy:

Elastic Potential Energy = (1/2) * k * x^2
= (1/2) * (490 N/m) * (0.400 m)^2
= 39.2 Joules

So, the elastic potential energy of the drawn bow is 39.2 Joules.

(b) To find the speed of the arrow when it leaves the bow, we need to equate the elastic potential energy to the kinetic energy of the arrow:

Elastic Potential Energy = Kinetic Energy

The kinetic energy of the arrow is given by:

Kinetic Energy = (1/2) * m * v^2

Where:
m is the mass of the arrow
v is the speed of the arrow

Given:
m = 0.0300 kg

Equating the elastic potential energy to the kinetic energy:

(1/2) * k * x^2 = (1/2) * m * v^2

Substituting the given values:

(1/2) * (490 N/m) * (0.400 m)^2 = (1/2) * (0.0300 kg) * v^2

Simplifying:

(0.5) * (490 N/m) * (0.1600 m^2) = (0.5) * (0.0300 kg) * v^2

19.6 N * m = 0.015 kg * v^2

Solving for v^2:

v^2 = (19.6 N * m) / (0.015 kg)
v^2 = 1306.67 m^2/s^2

Taking the square root on both sides:

v = √(1306.67 m^2/s^2)
v ≈ 36.14 m/s

So, the speed of the arrow when it leaves the bow is approximately 36.14 m/s.

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

Elastic Potential Energy (PE) = (1/2) * k * x^2

Where:
k is the spring constant (given as 490 N/m)
x is the displacement of the bowstring (given as 0.400 m)

(a) Calculating the elastic potential energy:
PE = (1/2) * 490 N/m * (0.400 m)^2
PE = 0.5 * 490 N/m * 0.16 m^2
PE = 39.2 J

Therefore, the elastic potential energy of the drawn bow is 39.2 Joules.

(b) To find the speed of the arrow when it leaves the bow, we can use the principle of conservation of mechanical energy:

Initial Energy = Final Energy
Potential Energy (PE) = Kinetic Energy (KE)

The initial energy is the elastic potential energy of the drawn bow, which is 39.2 J.

The final energy is the kinetic energy of the arrow when it leaves the bow.

We can calculate the kinetic energy using the formula:

KE = (1/2) * m * v^2

Where:
m is the mass of the arrow (given as 0.0300 kg)
v is the velocity of the arrow (what we need to find)

Substituting the values:
39.2 J = (1/2) * 0.0300 kg * v^2
78.4 J = 0.0300 kg * v^2

Rearranging the equation:
v^2 = (78.4 J) / (0.0300 kg)
v^2 = 2613.33 m^2/s^2

Taking the square root of both sides:
v = sqrt(2613.33 m^2/s^2)
v = 51.12 m/s (rounded to two decimal places)

Therefore, the arrow is traveling at a speed of approximately 51.12 m/s when it leaves the bow.