A 0.077 kg arrow is fired horizontally. The bowstring exerts an average force of 80 N on the arrow over a distance of 0.80 m. With what speed does the arrow leave the bow?

work done on arrow = force* distance = 80*.8 = 64 Joules

that is now kinetic energy
(1/2) m v^2 = 64
I think you can solve for v

how do you solve for v ?

V = sqroot of (2 x 64) / m

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 its kinetic energy.

The work done on the arrow is given by the product of the force applied and the distance over which it is applied. In this case, the force exerted by the bowstring is 80 N, and the distance over which it acts is 0.80 m. So the work done on the arrow is:

Work = Force × Distance = 80 N × 0.80 m

Next, we can calculate the change in kinetic energy of the arrow. Initially, the arrow is at rest, so its initial kinetic energy is zero. Finally, when the arrow leaves the bow, it will have some final kinetic energy. The change in kinetic energy is given by:

Change in Kinetic Energy = Final Kinetic Energy - Initial Kinetic Energy

Since the arrow starts from rest, the change in kinetic energy is equal to the final kinetic energy. We can equate this change in kinetic energy to the work done on the arrow.

Change in Kinetic Energy = Work

Finally, we can use the formula for kinetic energy to find the speed at which the arrow leaves the bow. Kinetic energy (KE) is given by the equation:

KE = (1/2) × mass × velocity^2

Rearranging the equation, we get:

velocity = sqrt(2 × KE / mass)

Substituting the value of change in kinetic energy calculated from the work done, and the mass of the arrow (0.077 kg), we can calculate the velocity:

velocity = sqrt(2 × Work / mass)

Therefore, to find the speed at which the arrow leaves the bow, calculate the work done on the arrow, substitute it into the equation for velocity, and solve for the answer.