To find the final speed at which the arrow would leave the bow after the force exerted on it is doubled, we need to apply the concept of work-energy theorem.
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done on the arrow will be equal to the change in its kinetic energy.
Initially, the arrow is at rest, so its initial kinetic energy is zero. The final speed of the arrow is what we need to find.
Let's denote the initial speed of the arrow as v_1 and the final speed as v_2.
According to the work-energy theorem, the work done on the arrow is equal to the change in its kinetic energy. Mathematically, this can be written as:
Work = ฮKinetic Energy
F * d = (1/2) * m * (v_2^2 - v_1^2)
Where:
F is the average force exerted on the arrow by the bow,
d is the distance over which the arrow is accelerated,
m is the mass of the arrow,
v_2 is the final speed of the arrow, and
v_1 is the initial speed of the arrow.
In this case, we are given that the initial speed of the arrow (v_1) is 28.8 m/s. We also know that the force is increased by a factor of 2.
First, let's consider the case where the force is not doubled. We can calculate the initial kinetic energy as:
Initial Kinetic Energy = (1/2) * m * v_1^2
Since the arrow starts from rest, the initial kinetic energy is zero.
Now, let's consider the case where the force is doubled:
New Force = 2 * Original Force
The work done with the new force can be calculated as:
New Work = New Force * d
Since the force is doubled, we have:
New Work = 2 * Original Work
According to the work-energy theorem, the work done on the arrow is equal to the change in its kinetic energy. Therefore:
2 * Original Work = (1/2) * m * (v_2^2 - v_1^2)
Now we can solve for v_2:
v_2^2 = (2 * Original Work * 2) / m + v_1^2
v_2 = sqrt((4 * Original Work) / m + v_1^2)
Using this equation, we can find the final speed at which the arrow would leave the bow after the force exerted on it is doubled.