A 0.0730 kg arrow is fired horizontally. The bowstring exerts an average force of 60.4 N on the arrow over a distance of 0.894 m. With what speed does the arrow leave the bow?
Set the work done drawing the bowstring equal to the kinetic energy of the released arrow. Solve for V
To find the speed at which the arrow leaves the bow, we can use the principle of work and energy. The work done on an object is equal to the change in the object's kinetic energy.
Let's break down the problem step by step:
Step 1: Find the work done on the arrow. The work done is given by the formula:
Work = Force * Distance * cos(theta)
In this case, the force is 60.4 N and the distance is 0.894 m. Since the arrow is fired horizontally, the angle between the force and the direction of motion is 0 degrees. So, cos(0°) = 1. Therefore:
Work = 60.4 N * 0.894 m * cos(0°) = 53.962 N·m
Step 2: Use the work-energy theorem to find the change in kinetic energy. The work done on the arrow is equal to the change in kinetic energy. The change in kinetic energy is equal to the final kinetic energy minus the initial kinetic energy.
In this case, the arrow starts from rest, so its initial kinetic energy is zero. The final kinetic energy is given by the formula:
Final Kinetic Energy = (1/2) * mass * velocity^2
Substituting the given values, the equation becomes:
53.962 N·m = (1/2) * 0.0730 kg * velocity^2
Simplifying the equation:
107.924 N·m = 0.0730 kg * velocity^2
Step 3: Solve for velocity. Rearranging the equation:
velocity^2 = 107.924 N·m / 0.0730 kg
velocity^2 = 1479.6164 m^2/s^2
Taking the square root of both sides:
velocity = sqrt(1479.6164) m/s
velocity ≈ 38.43 m/s
Therefore, the arrow leaves the bow with a speed approximately equal to 38.43 m/s.