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
Let's consider the conservation of mechanical energy for the banana as it falls, assuming there is no air resistance (the system is closed). The mechanical energy is composed of both potential energy (PE) and kinetic energy (KE).
At the start, when the monkey drops the banana from the top of the cliff, the banana has maximum potential energy (since it is at its highest point) and zero kinetic energy (since it is not moving). As the banana falls, the potential energy converts into kinetic energy.
The potential energy is given by 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 is given by KE = 1/2 mv^2, where v is the velocity of the falling banana.
At the bottom of the cliff, just before impact, the potential energy of the banana is essentially zero (because h is zero), and all the initial potential energy has been converted into kinetic energy. Therefore, the point where the potential energy and kinetic energy are equal must be somewhere between the top and the bottom.
The energies will be equal when the banana has fallen through half of the total height of the cliff, assuming no energy is lost to air resistance. This is because:
- At the top, PE is maximized, KE is zero.
- At the bottom, KE is maximized (all the initial PE has been converted to KE), PE is zero.
- At the halfway point, the banana has lost half of its initial potential energy (because it has fallen through half the height), and that amount of potential energy has been converted into kinetic energy.
So mathematically speaking:
PE_initial = KE_bottom,
which also equals:
1/2 PE_initial = 1/2 KE_bottom.
Since the potential and kinetic energies decrease and increase linearly respectively with height in this scenario, the halfway point is where PE = KE for the banana as it falls off the cliff.