The work-energy theorem relates work to kinetic energy. If a particle changes speed from v0 to v, what is the net work done on the particle?

The work-energy theorem relates work to kinetic energy. If a particle changes speed from v0 to v, what is the net work done on the particle?

• A. W_net = (1/2)m(v^2 - v0^2)
• B. W_net = (1/2)m(v0^2 - v^2)
• C. W_net = m(v - v0)
• D. W_net = m g h

Answer: (A)

The main idea is the work-energy principle: the net work done on a particle equals its change in kinetic energy. If the speed goes from v0 to v, the kinetic energy changes from (1/2) m v0^2 to (1/2) m v^2. So the net work is ΔK = (1/2) m v^2 − (1/2) m v0^2, which can be written as (1/2) m (v^2 − v0^2). This matches the given expression. The other forms don’t fit because one would yield the negative of the actual work, or imply a linear dependence on speed rather than on speed squared, or correspond to a specific gravitational potential-energy change that isn’t stated here.

The main idea is the work-energy principle: the net work done on a particle equals its change in kinetic energy. If the speed goes from v0 to v, the kinetic energy changes from (1/2) m v0^2 to (1/2) m v^2. So the net work is ΔK = (1/2) m v^2 − (1/2) m v0^2, which can be written as (1/2) m (v^2 − v0^2). This matches the given expression.

The other forms don’t fit because one would yield the negative of the actual work, or imply a linear dependence on speed rather than on speed squared, or correspond to a specific gravitational potential-energy change that isn’t stated here.