An arrow, starting from rest, leaves the bow with a speed of 28.8 m/s. If the average force exerted on the arrow by the bow was increased 2 times and the arrow was accelerated over the same distance, then with what speed would the arrow leave the bow?
Use the classical mechanic equations of force, displacement, and velocity under constant acceleration from rest.
(1)... F = m a
(2)... s = a t^2 /2
(3)... v = a t
From (1), doubling the force will double the acceleration (for the same mass).
From (2), doubling the acceleration while keeping the distance unchanged requires dividing time taken by the square root of 2.
From (3), doubling the acceleration and dividing time by the square root of 2, does ... what to the velocity?
To determine the final speed of the arrow when the average force exerted on it is increased by 2 times, we can use the principle of conservation of mechanical energy.
The initial speed of the arrow is given as 28.8 m/s, and since it starts from rest, we can take the initial kinetic energy as zero. Hence, the initial mechanical energy (Ei) is zero.
When the arrow leaves the bow, it will have some final mechanical energy (Ef), which is a combination of its final kinetic energy and potential energy.
According to the principle of conservation of mechanical energy, the change in the mechanical energy of the arrow is equal to the work done on it by the increased average force exerted on it.
Since the arrow is accelerated over the same distance, the work done on it is proportional to the average force applied.
Let's assume the increased average force is F1, and the final speed of the arrow is V1.
Using the work-energy theorem, we can write:
Work (W) = Change in Mechanical Energy (Ef - Ei)
= Increase in Kinetic Energy
= (1/2)m(V1^2 - 0)
= (1/2)mV1^2
The work done on the arrow is also given as the product of the distance (d) and the force (F). So we can write:
W = F1 * d
Since the average force is increased by 2 times, the new average force (F1) is 2 times the original average force. So we can write:
W = (2F) * d
Equating both expressions for work, we have:
(1/2)mV1^2 = (2F) * d
Simplifying the equation, we can write:
V1^2 = (4F * d) / m
Since the mass (m) and the distance (d) remain constant, and the force (F1) is 2 times the original force (F), we can write:
V1^2 = (4 * 2F * d) / m
V1^2 = (8F * d) / m
V1 = √((8F * d) / m)
Therefore, the arrow would leave the bow with a speed of √((8F * d) / m) when the average force exerted on the arrow is increased 2 times and it is accelerated over the same distance.
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 = ΔKinetic 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.