# Potential energy in a falling tree?

by kirderf
Tags: energy, falling, potential, tree
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,683 It doesn't take energy to stop a tree from falling, it takes force. Those are very different things. Probably the simplest way to find the gravitational potential energy in a tree, before it falls, relative to the ground, is to imagine it sliced at a great number of horizontal planes, each at a very small distance below the next. Let "A(z)" be the total area of the slice at height z. Its volume is "A(z)dz" where "dz" is the thickness of the slice and its mass is $\rho A(z)dz$ where $\rho$ is the density of tree. The potential energy of each slice is then $"mgh"= mgz= \rho A(z) z dz$. Add that up for all the slices. "In the limit", as we take more and more slices, and each slice thinner and thinner, it becomes the integral $mg\rho \int A(z)z dz$ where z goes from the bottom to the top of the tree. Notice that I said this is the potential energy of the tree before it falls. As the tree is falling, which was your question, the potential energy is continuously changing as the height of the various parts of the tree is continuously changing, some of the potential energy changing into kinetic energy. After the tree has fallen, its potential energy has changed, first to the kinetic energy of the tree, then to kinetic energy in shock waves in the tree and ground, and, eventually, to heat energy in the tree and the ground.