A 75 g arrow is fired horizontally. The bow string exerts an average of 65 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the bow string?

See 10:24 PM post.

To find the speed at which the arrow leaves the bow string, we can use the 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 by the bow string on the arrow is equal to the change in the arrow's kinetic energy.

The work done by a force can be calculated using the equation: work = force × distance × cos(theta), where theta is the angle between the direction of the force and the direction of the displacement.

In this problem, the force exerted by the bow string is 65 N, the distance over which the force is applied is 0.90 m, and the angle between the force and the displacement is 0 degrees (since the force is applied horizontally). Therefore, the work done on the arrow is:

work = 65 N × 0.90 m × cos(0°) = 58.5 J

According to the work-energy theorem, this work is equal to the change in the arrow's kinetic energy. Since the arrow starts from rest, its initial kinetic energy is zero. Therefore, the change in kinetic energy is equal to the final kinetic energy.

The equation for the kinetic energy of an object is: kinetic energy = 0.5 × mass × velocity^2

In this problem, the mass of the arrow is 75 g, which is equivalent to 0.075 kg. Let's assume the velocity at which the arrow leaves the bow string is v.

Therefore, the change in kinetic energy is:

58.5 J = 0.5 × 0.075 kg × v^2

Now we can solve this equation for v:

v^2 = (2 × 58.5 J) / (0.075 kg)
v^2 = 1560 J / 0.075 kg
v^2 = 20800 m^2/s^2

Taking the square root of both sides of the equation, we find:

v ≈ √20800 m^2/s^2
v ≈ 144 m/s

Therefore, the arrow leaves the bow string with a speed of approximately 144 m/s.