CrazyNeutrino said:So the proof would be...
mgy(b)+KE(b)=mgy(a)+KE(a)
That is: U(b)+K(b)=U(a)+K(a)
So U(b)-U(a)=K(a)-K(b)=-(K(b)-K(a)
So U(b)-U(a)=-W ( By work energy theorem )
Therefore:
ΔU=-W
Is this proof correct?
Delta PE is considered negative work because it represents the change in potential energy of a system as it moves from a higher potential energy state to a lower potential energy state. This change in potential energy is always negative because the system loses energy as it moves.
The relationship between delta PE and work is that delta PE is equal to the negative of the work done on the system. This means that the change in potential energy of a system is equal to the negative of the work done on the system.
The negative work of delta PE is derived using the formula W = F * d * cos(theta), where W represents work, F represents force, d represents displacement, and theta represents the angle between the force and displacement vectors. By substituting the formula for potential energy, PE = mgh, and rearranging the equation, we can derive the negative work of delta PE.
No, delta PE cannot be positive. This is because potential energy is always considered to be a measure of the system's energy relative to a reference point. Therefore, as the system moves from a higher potential energy state to a lower potential energy state, the change in potential energy will always be negative.
Understanding the relationship and derivation of delta PE can help in scientific research by providing a better understanding of how energy is transferred and transformed in various systems. This knowledge can be applied in fields such as physics, engineering, and environmental science to analyze and predict the behavior of different systems.