Compute work done by a force, including at an angle, and apply the work-energy theorem to relate net work to a change in kinetic energy.
In everyday speech "work" means effort, but in physics it has an exact meaning: work is done on an object only when a force causes it to move some distance, and it is computed as W = F*d*cos(theta), where d is the displacement and theta is the angle between the force and that displacement. Holding a heavy box perfectly still, no matter how tiring, does zero physics-work, because d=0. Work is measured in joules (J), where 1 J = 1 N*m. When a force pushes exactly along the direction of motion (theta=0), cos(theta)=1 and the formula simplifies to the familiar W=F*d; when a force is perpendicular to the motion (theta=90), cos(90)=0 and that force does no work at all, no matter how large it is.
The work-energy theorem ties this directly back to motion: the NET work done on an object (by every force combined) equals its change in kinetic energy, W_net = KE_final - KE_initial. This is a genuinely different and often faster way to solve motion problems than the kinematic equations from a few weeks ago -- instead of tracking time step by step, energy bookkeeping jumps straight from a starting speed to an ending speed using only forces and distances.
Mastery looks like: They correctly compute work done by a force (including at an angle) and apply the work-energy theorem to connect net work to a change in kinetic energy.
Common stumbles: Forgetting the cos(theta) factor when a force is not aligned with the motion; confusing "effort" or time spent with physics-work, which depends only on force and displacement.