Question

an 0.05kg arrow in a bow has 50j of potential energy when the archer pulls the string all the way back and holds it steady while aiming at the target. how much kinetic energy will it have when it is shot from the bow and the string returns to its unstretched position?
0j
102j
25j
50j

Explanation:

Step1: Recall Energy Conservation

In a bow - arrow system (assuming no non - conservative forces like air resistance or energy loss in the bow), the total mechanical energy is conserved. The mechanical energy \(E\) is the sum of potential energy (\(PE\)) and kinetic energy (\(KE\)), i.e., \(E = PE+KE\).

When the archer holds the string back, the arrow has only potential energy (since it is at rest, \(KE = 0\)) and \(PE=50\space J\), so the total mechanical energy \(E = 50\space J+0\space J = 50\space J\).

Step2: Analyze Energy at Unstretched Position

When the string returns to its unstretched position, the potential energy stored in the bow (elastic potential energy) is zero (\(PE = 0\)). According to the law of conservation of mechanical energy, the total mechanical energy remains the same. So, \(E=PE + KE\). Substituting \(E = 50\space J\) and \(PE = 0\space J\) into the equation, we get \(50\space J=0\space J + KE\). Solving for \(KE\), we find that \(KE = 50\space J\).

Answer:

D. 50J (assuming the last option is D. 50J, if the options are labeled as: A. 0J, B. 102J, C. 25J, D. 50J)