Question

at what position does the rock have the greatest kinetic energy?

Explanation:

Step1: Recall the formula for kinetic energy

The formula for kinetic energy is \( KE=\frac{1}{2}mv^{2}\), where \(m\) is the mass of the object and \(v\) is the velocity of the object. Kinetic energy depends on the velocity of the object.

Step2: Analyze the velocity at each position

• At position \(A\) and \(B\): The rock is being lifted or is at rest (assuming no motion other than being held). So, the velocity \(v = 0\) (or very low if just starting to lift). So, \(KE=\frac{1}{2}m\times0^{2}=0\) (or very small).
• At position \(C\): The rock is just about to be dropped. Its velocity \(v = 0\) (initial velocity of fall). So, \(KE=\frac{1}{2}m\times0^{2}=0\).
• At position \(D\): The rock is in the process of falling. It has some velocity \(v>0\), but not the maximum.
• At position \(E\): As the rock falls, it accelerates due to gravity (\(F = mg\), \(a = g\) in the absence of air - resistance). Just before hitting the ground (assuming the ground is the lowest point in the context of the problem), the rock has the maximum velocity \(v_{max}\). Using \(KE=\frac{1}{2}mv^{2}\), with \(v = v_{max}\), the kinetic energy \(KE\) is maximum.

Answer:

E