An arrow is placed on a bow, the bowstring is pulled back, and the arrow is shot straight up into the air; the arrow then comes back down and sticks into the ground. Describe all of the changes in work and energy that occur.

There are three main forces acting on the arrow during its motion: 1. Gravitational force (weight): it acts downward, opposite to the upward motion of the arrow, and its work will be taken into account when the arrow is gaining altitude. When the arrow falls back to the ground, the force is in the same direction as the movement, and its work will be considered in that phase. 2. Tensile force in the bowstring: it acts during the initial phase, as the bowstring is pulled back and then released. Its work should be considered when the arrow is shot into the air. 3. Resistive force (air resistance): it opposes the arrow's motion during ascent and descent. The work of this force should be taken into account in both phases, although it may be negligible due to the relatively small size and shape of the arrow.

Initially, the arrow has potential energy due to its drawn position on the bow string. This energy is then converted into kinetic energy when the arrow is released, and the bowstring work can be considered together with potential energy form. Therefore, at the moment of release, the arrow has only kinetic energy: \(E_k= \frac{1}{2}mv^2\) where \(m\) is the mass of the arrow, and \(v\) is the initial speed.

As the arrow travels upward, the gravitational force does negative work on it due to the opposite direction to the arrow's motion. The energy conversion between kinetic and potential energy takes place. The kinetic energy decreases while the potential energy increases: \(\Delta E_p = mgh\) where \(m\) is the mass of the arrow, \(g\) is the acceleration due to gravity, and \(h\) is the height reached. At the highest point, the arrow has no kinetic energy (because its velocity is momentarily 0), and all of the initial energy has been converted into potential energy: \(E_{k,highest} = 0\) \(E_{p,highest} = mgh_{highest}\)

When the arrow starts to descend, gravitational force now does positive work on the arrow, converting potential energy back into kinetic energy. As the arrow falls and its altitude decreases, its potential energy decreases while its kinetic energy increases: \(\Delta E_p = -mgh\) Upon reaching the ground, the arrow's potential energy is zero and its kinetic energy is: \(E_{k,impact} = \frac{1}{2}mv_{impact}^2\) If air resistance is negligible, the arrow will reach the ground with the same kinetic energy it had initially and stick into the ground. The total change in work and energy during the arrow's motion is thus related to the conversion between kinetic and potential energy during the ascent and descent.