if a monkey drops a banana off a cliff, where are the potential and kinetic energies of the banana equal?
at the top
at the bottom
a quarter of the way down
half way down
The potential and kinetic energies of the banana would be equal halfway down the cliff. Potential energy is maximized at the top of the cliff where the banana has the greatest height, and kinetic energy is maximized just before it hits the ground, where its velocity is greatest.
As the banana falls, its potential energy is converted into kinetic energy. The potential energy (PE) is given by the equation PE = mgh, where m is the mass of the banana, g is the acceleration due to gravity, and h is the height above the ground. The kinetic energy (KE) is given by KE = 1/2 mv², where v is the velocity of the banana.
At the halfway point, the banana has lost half of its potential energy and gained an equivalent amount of kinetic energy. This is because the conversion between kinetic and potential energy is a trade-off; as one decreases, the other increases proportionally (ignoring air resistance and other non-conservative forces). Since energy is conserved in a closed system, the points where the potential and kinetic energies are equal represents a 50-50 split between the two forms of energy.