The discussion revolves around the concept of work in physics, specifically the relationship between work, kinetic energy, and potential energy. Participants explore the definitions and implications of work as described in textbooks, questioning whether work should be defined solely as the change in kinetic energy or if it encompasses changes in potential energy as well.
The discussion is active, with participants providing insights and raising questions about the definitions and principles involved. Some participants offer clarifications regarding the work-energy principle and the role of net work, indicating a productive exploration of the topic without reaching a consensus.
Participants note potential ambiguities in textbook explanations and the need for clarity regarding the definitions of work, kinetic energy, and potential energy. There is also mention of specific scenarios, such as lifting objects at constant velocity, which complicate the understanding of work done in different contexts.
guitarphysics said:Oh, net work :). That's the big thing here! So look, when you lift something upwards, you're doing work on it, right? BUT it's not changing in kinetic energy! How? Because gravity is doing work that is equal to yours but opposite in sign, so there is zero NET work done on the object, thus its kinetic energy won't change ;).
Doc Al said:Hyperphysics is not defining work via the change in kinetic energy, they are describing the work-energy principle. And, as guitarphysics points out, it is the net work on a particle (including all forces acting) that gives the change in kinetic energy.
It's the net work that gives the change in KE. Of course, if only one force acts then the work it does will equal the change in KE, since that is the net force.dragon-kazooie said:So, you're saying that work is equal to Δk or -ΔU, but net work is only equal to Δk? (Because in the crane example, there's still a change in potential energy, even though there's no change in kinetic energy and no net work.) That seems really weird.