Question

determine how potential and kinetic energy changes at each position of the pendulum as the ball swings from a to e. position a: position b: position c: position d: position e:

Explanation:

• Position A: The ball is at the highest point (maximum height). Using the formula for gravitational potential energy \(U = mgh\) (where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is height), since \(h\) is maximum, potential energy is maximum. At the extreme - position (A), the velocity \(v = 0\). Using the formula for kinetic energy \(K=\frac{1}{2}mv^{2}\), kinetic energy is minimum (zero).
• Position B: As the ball moves from A to B, it loses height (\(h\) decreases, so potential energy \(U=mgh\) decreases) and gains velocity (\(v\) increases, so kinetic energy \(K = \frac{1}{2}mv^{2}\) increases).
• Position C: The ball is at the lowest point (minimum height). So, potential energy \(U=mgh\) is minimum. The velocity \(v\) is maximum at this point (using conservation of mechanical energy \(E=U + K\), as \(U\) is minimum, \(K=\frac{1}{2}mv^{2}\) is maximum).
• Position D: As the ball moves from C to D (similar to A - B but in the opposite direction in terms of motion), it gains height (\(h\) increases, so \(U=mgh\) increases) and loses velocity (\(v\) decreases, so \(K=\frac{1}{2}mv^{2}\) decreases).
• Position E: Similar to position A. The ball is at a height (same as A in an ideal pendulum with no energy loss). So, potential energy \(U = mgh\) is maximum and kinetic energy \(K=\frac{1}{2}mv^{2}\) is minimum (zero)

Answer:

• Position A: Maximum potential energy, minimum kinetic energy
• Position B: Decreasing potential energy, increasing kinetic energy
• Position C: Minimum potential energy, maximum kinetic energy
• Position D: Increasing potential energy, decreasing kinetic energy
• Position E: Maximum potential energy, minimum kinetic energy