For a particle moving under forces, which statement expresses the work-energy theorem?

• A. W_total = ΔU
• B. W_total = m g h
• C. (1/2) m v^2
• D. W_total = ΔK = (1/2) m v^2 − (1/2) m v0^2

Answer: (D)

The work-energy theorem says that the total work done on a particle equals the change in its kinetic energy. The kinetic energy is (1/2) m v^2, so going from initial speed v0 to final speed v the change is (1/2) m v^2 − (1/2) m v0^2. Therefore the total work done by all forces is W_total = ΔK = (1/2) m v^2 − (1/2) m v0^2. This is the general statement because it accounts for any forces doing work, not just gravity or a potential-energy change. The other expressions don’t capture the whole picture: ΔU would describe only potential-energy changes for conservative forces, m g h is the work done by gravity in a vertical move, and (1/2) m v^2 alone is the final kinetic energy, not the change.