Potential energy in a falling tree?

  1. Hi,
    My first post here, hope it goes well.
    I wonder if someone of you skilled persons could help me and my friend calculate the energy in a falling tree.
    I´m no good at maths neither good in english but I will try to explain what I´m after.
    If you cut down a tree that is 25m heigh and weighs approx 700kg. How much energy does it produce when falling to ground?
    Or, if different, if you trying to hold back the energy 1m up on the stem, how much energy must you use to stop it?

    Hope someine can help me this, thanks in advance.
    Note, it is not homework.
     
  2. jcsd
  3. Simon Bridge

    Simon Bridge 14,657
    Science Advisor
    Homework Helper
    Gold Member

    Welcome to PF;
    Taking the last part first: it takes no energy to hold back the tree. This is because nothing is moving.
    A person trying that on a substantial tree may be under considerable strain though.

    The change in gravitational potential energy, when something falls close to the earth's surface, is equal to mgh
    Here g is the acceleration of gravity, about 9.8m/s/s; m is the total mass of the object; and h is the distance traveled by the object's center of mass.
     
  4. HallsofIvy

    HallsofIvy 40,307
    Staff Emeritus
    Science Advisor

    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 [itex]\rho A(z)dz[/itex] where [itex]\rho[/itex] is the density of tree. The potential energy of each slice is then [itex]"mgh"= mgz= \rho A(z) z dz[/itex]. 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 [itex]mg\rho \int A(z)z dz[/itex] 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.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted