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Grade 11-12 (age 16-18)

Work & Work–Energy Theorem

Work & Work–Energy Theorem

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
8 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
Wnet = ΔEk = 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.
1
Find the work done by the constant force.
W = F s = 4 × 4 = 16
2
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.
1
Work = area under the F–x graph. Rectangle (0–4 m): 10 × 4 = 40 J.
2
Triangle (4–6 m): ½ × 2 × 10 = 10 J. Total = 50 J → ①, ② wrong; ④ wrong (a force still does work).
3
By the work–energy theorem ΔEk = 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
③Work of a varying force = area under the F–x graph
④The change in kinetic energy is independent of the initial speed — set by the net work only
⑤Friction does negative work, reducing kinetic energy
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