An archer puts a 0.26 kg arrow to the bowstring. An average force of 184.5 N is exerted
to draw the string back 1.26 m.
The acceleration of gravity is 9.8 m/s
2
.
Assuming no frictional loss, with what
speed does the arrow leave the bow?
Answer in units of m/s
Well, let's see how this archery problem shoots itself in the foot with some physics humor!
To find the speed at which the arrow leaves the bow, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the bowstring is converted into kinetic energy as the arrow is released.
First, we need to calculate the potential energy stored in the bowstring. We can do this using the formula:
Potential Energy (PE) = mgh,
where m is the mass of the arrow (0.26 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height difference (1.26 m) that the string is pulled back.
PE = (0.26 kg)(9.8 m/s^2)(1.26 m)
Now, let's convert this potential energy to kinetic energy using the equation:
Kinetic Energy (KE) = 0.5mv^2,
where v is the velocity of the arrow.
So, we set the potential energy equal to the kinetic energy:
PE = KE,
(mgh) = (0.5mv^2).
Now, we can solve for v:
v^2 = (2gh),
v = sqrt(2gh).
Substituting the values:
v = sqrt(2 * 9.8 m/s^2 * 1.26 m).
Calculating that:
v ≈ 5.01 m/s.
So, the arrow leaves the bow with a speed of approximately 5.01 m/s. That's one fast arrow flying through the air! Keep shooting for the stars, my friend!
To find the speed at which the arrow leaves the bow, we can use the principle of conservation of mechanical energy. Assuming no frictional loss, the total mechanical energy remains constant throughout the process.
The initial potential energy (due to the arrow being raised) is converted into kinetic energy (as the arrow leaves the bow).
The potential energy can be calculated using the formula:
Potential energy (PE) = mass (m) * acceleration due to gravity (g) * height (h)
In this case, the height is given as 1.26 m and the acceleration due to gravity is 9.8 m/s^2, so the potential energy becomes:
PE = 0.26 kg * 9.8 m/s^2 * 1.26 m
Next, we can calculate the kinetic energy using the formula:
Kinetic energy (KE) = 0.5 * mass (m) * velocity squared (v^2)
We can equate the potential energy to the kinetic energy:
PE = KE
0.26 kg * 9.8 m/s^2 * 1.26 m = 0.5 * 0.26 kg * v^2
Now we can solve for the velocity (v):
v^2 = (2 * PE) / m
v^2 = (2 * (0.26 kg * 9.8 m/s^2 * 1.26 m)) / 0.26 kg
Simplifying the equation:
v^2 = 2 * 9.8 m/s^2 * 1.26 m
v^2 = 24.552 m^2/s^2
Finally, taking the square root of both sides, we find the speed at which the arrow leaves the bow:
v = √(24.552 m^2/s^2)
Calculating this:
v ≈ 4.96 m/s
Therefore, the arrow leaves the bow with a speed of approximately 4.96 m/s.
To find the speed at which the arrow leaves the bow, we can use the principle of conservation of mechanical energy. The mechanical energy of the arrow consists of its potential energy when the string is drawn back and its kinetic energy when it leaves the bow.
1. Calculate the potential energy of the arrow when the string is drawn back:
The potential energy of an object can be calculated using the formula:
Potential Energy (PE) = mass (m) x gravity (g) x height (h)
In this case, the mass of the arrow is 0.26 kg and the height is the distance the string is drawn back, which is 1.26 m. The acceleration due to gravity is given as 9.8 m/s^2.
PE = 0.26 kg x 9.8 m/s^2 x 1.26 m
2. Calculate the kinetic energy of the arrow when it leaves the bow:
The kinetic energy of an object can be calculated using the formula:
Kinetic Energy (KE) = 0.5 x mass (m) x velocity^2 (v^2)
In this case, we need to find the velocity at which the arrow leaves the bow. Since no frictional losses are assumed, the potential energy is converted entirely into kinetic energy.
PE = KE
0.26 kg x 9.8 m/s^2 x 1.26 m = 0.5 x 0.26 kg x v^2
Solve for v^2:
0.26 kg x 9.8 m/s^2 x 1.26 m = 0.5 x 0.26 kg x v^2
v^2 = (0.26 kg x 9.8 m/s^2 x 1.26 m) / (0.5 x 0.26 kg)
v^2 = 25.852
v ≈ √25.852 ≈ 5.08 m/s
Therefore, the speed at which the arrow leaves the bow is approximately 5.08 m/s.
(Average force) x (draw back distance) = work done on bow
Set that equal to the arrow's kinetic energy in flight. Solve for V
You will not need to use the acceleration of gravity.
(M/2)*V^2 = 184.5*1.26 = 232.5 J
V = sqrt(2*232.5/0.26)= 42.3 m/s