Energy Transformation equation for a free fall

AI Thread Summary
The energy transformation equation for a free-falling object is expressed as Total Energy = Potential gravitational energy + Kinetic energy, or Et = mgh + 1/2 mv^2. In this context, total energy is conserved, meaning mgh + 1/2 mv^2 equals a constant value, C. To demonstrate that energy remains constant throughout the fall, one can substitute the equations for height and velocity over time into the energy equation. This confirms that energy transformation occurs seamlessly between potential and kinetic forms during free fall. Understanding these principles is crucial for analyzing energy conservation in physics.
jman1114
Messages
3
Reaction score
0
One of the discussion questions in a lab I did which was basically a free falling object where the kinetic and potential energies were measured. It asks what is the energy transformation equation for a free fall (1-step transformation).
Is it just:
Total Energy = Potential gravitational energy + Kinetic energy
Et= mgh + mv2/2

Then just cancel out the mass or am I going in the total wrong direction with this?
 
Physics news on Phys.org
Total energy is conserved, which means that mgh + 1/2 mv^2 = C, where C is a constant. Use the equations for h(t) and v(t) for a free fall and plug them into the energy equation to convince yourself that energy is constant for every moment t.
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top