An archer puts a 0.39 kg arrow to the bowstring.
An average force of 196.3 N is exerted
to draw the string back 1.28 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
V^2 = Vo^2 + 2a*d
Vo = 0
a = F/m = 196.3/0.39 =
d = 1.28 m.
Solve for V.
To find the speed at which the arrow leaves the bow, we can make use of the principle of conservation of energy. We can analyze the situation in terms of potential energy and kinetic energy.
First, let's calculate the potential energy of the bowstring when it is drawn back. The potential energy equation is given by:
Potential energy (PE) = mass (m) * acceleration due to gravity (g) * height (h)
Here, the mass of the arrow is given as 0.39 kg, the acceleration due to gravity is 9.8 m/s^2, and the height is the distance the bowstring is drawn back, which is 1.28 m. Plugging in these values, we get:
PE = 0.39 kg * 9.8 m/s^2 * 1.28 m
PE = 4.882 N*m
Next, let's calculate the kinetic energy of the arrow when it leaves the bow. The kinetic energy equation is given by:
Kinetic energy (KE) = 0.5 * mass (m) * velocity squared (v^2)
We want to find the velocity (v), so we rearrange the equation as:
v^2 = (2 * KE) / m
We know the mass of the arrow is 0.39 kg, and the kinetic energy is equal to the potential energy when the arrow is released from the bowstring. Therefore, the kinetic energy is also 4.882 N*m. Plugging in these values, we get:
v^2 = (2 * 4.882 N*m) / 0.39 kg
v^2 = 25.038 m^2/s^2
To find the velocity, we take the square root of both sides:
v = √(25.038 m^2/s^2)
v ≈ 5.004 m/s
Therefore, the speed at which the arrow leaves the bow is approximately 5.004 m/s.