A 0.052-kg arrow is fired horizontally. The bowstring exerts an average force of 80 N on the arrow over a distance of 0.75 m.
With what speed does the arrow leave the bow?
The average force multiplied by the 0.75 m distance is the work done on the arrow, and equals the kinetic energy of the arrow when it leaves the bow.
Therefore 80 x 0.75 = 60 J = (1/2) m V^2
You know what the arrow's mass (m) is; so solve for V
10.95 m/s
To find the speed at which the arrow leaves the bow, we can use the principle of work and energy.
First, let's calculate the work done on the arrow by the bowstring. The work done is equal to the force applied multiplied by the distance over which the force is applied:
Work = Force x Distance
Given that the force applied by the bowstring is 80 N and the distance over which the force is applied is 0.75 m, we can substitute these values into the equation:
Work = 80 N x 0.75 m = 60 J
The work done on the arrow is 60 joules.
Now, let's use the concept of kinetic energy to find the speed of the arrow. The kinetic energy of an object is given by the equation:
Kinetic Energy = (1/2) x mass x velocity^2
In this case, the arrow's mass is 0.052 kg. The kinetic energy, which is equal to the work done on the arrow, is 60 J. We can substitute these values into the equation:
60 J = (1/2) x 0.052 kg x velocity^2
Now, let's solve for the velocity:
(1/2) x 0.052 kg x velocity^2 = 60 J
0.026 kg x velocity^2 = 60 J
velocity^2 = 60 J / 0.026 kg
velocity^2 = 2307.69 m^2/s^2
To find the velocity, we take the square root of both sides:
velocity = √(2307.69 m^2/s^2)
velocity ≈ 48.04 m/s
Therefore, the arrow leaves the bow with a speed of approximately 48.04 m/s.