The discussion focuses on the relationship between gravitational energy and kinetic energy for an object in free fall. When an object is dropped from a height, h, its gravitational energy is calculated as E = mgh, while its kinetic energy upon impact is given by E = (1/2)mv². The principle of conservation of energy asserts that these two forms of energy are equal at the moment of impact, leading to the conclusion that the speed of the object can be expressed as v = √(2gh). This demonstrates that kinetic energy does depend on the height from which the object is dropped, as it is derived from gravitational potential energy.
PREREQUISITESStudents of physics, educators teaching mechanics, and anyone interested in the principles of energy conservation and motion dynamics.
How does the h cancel out?Shyan said:It does! Its equal to mgh which is a function of h!
It doesn't. ## E_k=\frac 1 2 m v^2 ## is the definition of kinetic energy of a particle with mass m and speed v regardless of what forces are applied to it. Any particle with mass m and speed v, has the kinetic energy ## E_k ##. Now if this particle acquired its kinetic energy by falling from a height h under constant gravitational acceleration g, then the numerical value of ## mgh ## and ## \frac 1 2 m v^2 ## will be equal which means the speed of the particle is ## v=\sqrt{2gh} ##.bay said:How does the h cancel out?