Final answer:
The potential energy of the ball when dropped is 5 J, which will be fully converted to kinetic energy just before it hits the ground, also totaling 5 J, if we ignore air resistance.
Step-by-step explanation:
If a ball is dropped and has a potential energy (P.E) of 5 J, by the principle of conservation of energy, this potential energy will be converted into kinetic energy (K.E) just before it hits the ground, assuming air resistance is negligible. Therefore, the kinetic energy of the ball just as it hits the ground would also be 5 J, because the potential energy has been completely transformed into kinetic energy.
The principle of conservation of energy states that energy cannot be created or destroyed, but can only be converted from one form to another. In a falling object, potential energy decreases as height decreases, and an equivalent amount of kinetic energy is gained as the object's velocity increases.
An example to illustrate this concept can be seen when a 10 kg bowling ball falls from a height of 5 m. If we use the approximation that g (acceleration due to gravity) is about 10 m/s², we first calculate the gravitational potential energy using the formula P.E = mass * g * height. So, the potential energy would be P.E = 10 kg * 10 m/s² * 5 m = 500 J. As the bowling ball reaches the ground, all 500 J of potential energy would have been converted into kinetic energy, and the ball's velocity can be calculated using the kinetic energy formula K.E = 0.5 * mass * velocity².