Potential energy in a falling tree?

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SUMMARY

The potential energy of a falling tree can be calculated using the formula for gravitational potential energy, expressed as mgh, where m is the mass (700 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (25 m). The energy produced when the tree falls is equal to 171,500 joules. It is important to note that holding back the tree does not require energy, but rather force, as the tree is not in motion until it falls. The potential energy transforms into kinetic energy during the fall and eventually dissipates as heat energy upon impact.

PREREQUISITES
  • Understanding of gravitational potential energy and its formula (mgh)
  • Basic knowledge of physics concepts such as mass, height, and gravity
  • Familiarity with calculus for integrating functions (e.g., ∫ A(z)z dz)
  • Concept of energy transformation (potential to kinetic energy)
NEXT STEPS
  • Study the principles of energy conservation in physics
  • Learn about the integration of functions in calculus
  • Explore the relationship between potential and kinetic energy in falling objects
  • Investigate the effects of impact energy on different materials
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Physics students, educators, engineers, and anyone interested in understanding the mechanics of falling objects and energy transformations.

kirderf
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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.
 
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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.
 

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