Potential energy in a falling tree?

AI Thread Summary
To calculate the energy produced by a falling tree, the change in gravitational potential energy can be determined using the formula mgh, where m is the mass (700 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height (25 m). It is important to note that holding back the tree does not require energy, as it is a static force, but rather a significant force is needed to counteract its weight. The potential energy of the tree changes continuously as it falls, converting into kinetic energy and eventually into other forms of energy such as heat upon impact. The calculation involves integrating the potential energy of the tree's slices at various heights before it falls. Understanding these energy transformations is crucial in analyzing the dynamics of a falling tree.
kirderf
Messages
1
Reaction score
0
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.
 
Mathematics news on Phys.org
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.
 
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.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top