Work & Work–Energy Theorem

Work is force times distance moved along the force, and it is zero when force is perpendicular to the motion. Net work equals the change in kinetic energy, and for a varying force the work is the area under an F-x graph. Without friction that work all becomes kinetic energy and sets the final speed. Slide the applied force here to see how the graph area and final speed change.

A force acting over a distance does work

💪 What is work?

①Work = force × distance moved along the force direction

②If the force is perpendicular to the motion, the work is zero (e.g. the centripetal force in uniform circular motion)

③The work done by the net force changes the object's kinetic energy

Change the force to see work and speed

Applied force F8 N

📊 The area IS the work

①For a constant force the F–x graph is a rectangle; area = F × s = work

②All the work becomes kinetic energy → ½mv² = W

③A larger force means a larger area (work) and a higher final speed

Work and kinetic energy in equations

Work

W = F s cosθ

θ = angle between force and motion; if perpendicular (90°), W=0

Work–energy theorem

W_net = ΔE_k = 12mv² − 12mv₀²

work by the net force = change in kinetic energy

Work done by a varying force

W = (area under the F–x graph)

even when the force is not constant, the area gives the work

From concept to problem

Example 1 — work and final speed

On a frictionless surface, a constant force of 4 N acts on a 2 kg object at rest, along the direction of motion, over 4 m. Find the work done and the final speed.

Find the work done by the constant force.

W = F s = 4 × 4 = 16

Use the work–energy theorem for the final speed (starts at rest).

16 = 12(2)v² → v² = 16 → v = 4

▸ work 16 J, final speed 4 m/s

All the net work becomes kinetic energy. With friction, that much leaves as heat and the speed is lower.

Exam Points

CSAT-style (adapted, Physics II)

On a frictionless line, the net force F on an object at rest, plotted against position x, is 10 N (constant) over 0–4 m and decreases linearly from 10 N to 0 over 4–6 m. Choose the correct statement.

①The net force does 40 J of work over 0–6 m.

②The net force does 60 J of work over 0–6 m.

③The net force does 50 J of work and the kinetic energy increases by 50 J.

④Over 4–6 m the force decreases, so it does no work.

⑤The change in kinetic energy can only be found if the initial speed is known.

▸ ③ The net force does 50 J of work and the kinetic energy increases by 50 J.

Work = area under the F–x graph. Rectangle (0–4 m): 10 × 4 = 40 J.

Triangle (4–6 m): ½ × 2 × 10 = 10 J. Total = 50 J → ①, ② wrong; ④ wrong (a force still does work).

By the work–energy theorem ΔE_k = W = 50 J; the change is independent of the initial speed → ⑤ wrong, ③ correct.

🎯 Exam Points

①Work W = Fs cosθ; if force ⊥ motion, W=0

②Work–energy theorem: net work = ΔE_k = ½mv²−½mv₀²