A 140 g arrow is shot vertically from a bow whose effective spring constant is 500 N/m. If the bow is drawn 60 cm before shooting the arrow,to what height does the arrow rise?
M g H = (1/2) k X^2
X = 0.60 m
M = 0.140 kg
k = 500 N/m
g = 9.8 m/s^2
Solve for H, in meters.
H = [kX^2/(2Mg)]
Well, shooting an arrow vertically can be quite the spring-loaded adventure! Let me calculate the height it rises for you.
First, let's convert the distance the bow is drawn from centimeters to meters. 60 cm is equal to 0.6 meters. Got it? Great!
Now, we can calculate the potential energy stored in the bow when it's drawn. The formula for potential energy is given by PE = (1/2)kx^2, where PE is the potential energy, k is the spring constant, and x is the distance the bow is drawn.
Substituting the values we have, PE = (1/2) * 500 N/m * (0.6 m)^2. Crunching the numbers, we get PE = 90 Joules.
But wait, we can't forget about the arrow! The potential energy stored in the bow is then transferred to the arrow as it is shot upwards. We can calculate the arrow's maximum height by using the equation for gravitational potential energy, mgh, where m is the mass of the arrow, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the maximum height.
Rearranging the equation, we have h = PE / (mg). Substituting the values, we get h = 90 J / (0.14 kg * 9.8 m/s^2). After some math, we find that the arrow rises to a height of approximately 64.3 meters.
So, this arrow reaches heights that would make even the tallest giraffe jealous! Keep aiming high!
To find the height to which the arrow rises, we can use conservation of mechanical energy.
The potential energy when the bowstring is drawn back is converted into the potential energy when the arrow is at its maximum height. The equation for the potential energy of a spring is given by:
U = (1/2)kx^2,
where U is the potential energy, k is the spring constant, and x is the displacement.
Given:
mass of the arrow (m) = 140 g = 0.14 kg
spring constant (k) = 500 N/m
displacement (x) = 60 cm = 0.6 m
The potential energy when the bowstring is drawn back is given by:
U_initial = (1/2)kx^2,
= (1/2)(500 N/m)(0.6 m)^2,
= (1/2)(500 N/m)(0.36 m^2),
= 90 J.
The potential energy at maximum height is equal to the gravitational potential energy:
U_final = mgh,
= (0.14 kg)(9.8 m/s^2)h,
= 1.372 h J.
Setting the initial and final potential energies equal to each other:
U_initial = U_final,
90 J = 1.372 h J.
Now solve for h:
h = 90 J / 1.372 J,
h ≈ 65.6 m.
Therefore, the arrow rises to a height of approximately 65.6 meters.
To find the height to which the arrow rises, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the bow is equal to the final potential energy of the arrow at its highest point.
1. Determine the initial potential energy:
The potential energy stored in the bow is given by the formula:
Potential energy (PE) = (1/2) * k * x^2
where k is the effective spring constant and x is the displacement of the bow (drawn length).
Given that k = 500 N/m and x = 0.6 m, we can calculate the initial potential energy.
PE_initial = (1/2) * 500 N/m * (0.6 m)^2
2. Determine the final potential energy:
The final potential energy of the arrow at its highest point (when its velocity becomes zero) can be calculated using the formula:
PE_final = m * g * h
where m is the mass of the arrow, g is the acceleration due to gravity (9.8 m/s^2), and h is the height to which the arrow rises.
Given that m = 0.14 kg, we can calculate the final potential energy.
PE_final = 0.14 kg * 9.8 m/s^2 * h
3. Equate the two potential energies:
Since the initial potential energy is equal to the final potential energy, we can equate them and solve for h.
(1/2) * 500 N/m * (0.6 m)^2 = 0.14 kg * 9.8 m/s^2 * h
Simplifying the equation:
150 N * 0.36 m^2 = 1.372 kg*m/s^2 * h
54 N*m = 1.372 kg*m/s^2 * h
Dividing both sides of the equation by 1.372 kg*m/s^2:
h = 54 N*m / (1.372 kg*m/s^2)
Finally, calculating h:
h ≈ 39.38 meters
Therefore, the arrow rises to a height of approximately 39.38 meters.