A \( 2.00-\mathrm{kg} \) ball has zero kinetic and potential energy. Ernie drops the ball into a 10.0 - m-deep well. Just before the ball hits the bottom, the sum of its kinetic and potential energy is: 1). zero. 2) 196 J . 3) -196 J . 4) 392 J . 5) -392 J .

To solve this problem, we need to consider the conservation of energy. The total energy of the ball remains constant throughout the process of falling into the well. Given: - Initial kinetic energy = 0 - Initial potential energy = 0 - Depth of the well = 10.0 m - Mass of the ball = 2.00 kg Let's denote the final kinetic energy as \( K_f \) and the final potential energy as \( U_f \). The total energy at the bottom of the well is the sum of the kinetic energy and the potential energy: \[ E_{\text{total}} = K_f + U_f \] The potential energy at the bottom of the well is given by: \[ U_f = mgh \] where: - \( m \) is the mass of the ball - \( g \) is the acceleration due to gravity (approximately 9.8 m/s^2) - \( h \) is the depth of the well Substitute the values into the equation for potential energy: \[ U_f = 2.00 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 10.0 \, \text{m} \] Calculate the potential energy: \[ U_f = 196 \, \text{J} \] Since the initial kinetic energy is zero, the total energy at the bottom of the well is equal to the potential energy: \[ E_{\text{total}} = U_f = 196 \, \text{J} \] Therefore, the sum of the kinetic and potential energy just before the ball hits the bottom is 196 J. The correct answer is 2) 196 J.