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Grade 11-12 (age 16-18)

high school physics Work & Work–Energy Theorem

Work is a force multiplied by the distance moved in the direction of that force, so when the force is perpendicular to the motion, the work is 0 no matter how hard you push. The work done by the net force equals the change in the object's kinetic energy, and when the force is not constant, the work is the area under the F-x graph. Without friction, all the work done becomes kinetic energy, so the amount of work fixes the final speed. In the picture, changing the size of the force changes both the area under the graph and the final speed of a 2 kg object that started at rest.

A Force Acting Over a Distance Does Work

💪 What Work Is
①Work is force × distance moved in the direction of the force
②When the force is perpendicular to the motion, the work is 0 (for example, the centripetal force in uniform circular motion)
③The kinetic energy of an object changes by exactly the work done by the net force

Work and Speed for Different Forces

8 N
📊 The Area Is the Work
①For a constant force the region under the F-x graph is a rectangle, and its area F × s is the work done
②Without friction, all the work becomes kinetic energy, so ½mv² = W
③Doubling the force doubles the work, but the final speed grows only by a factor of √2
④When the force changes with position, the region under the graph becomes a trapezoid or a triangle, and its area is still the work
⑤A force opposite to the motion (θ = 180°), like friction, makes W = Fs cosθ negative and reduces the kinetic energy

Work and Kinetic Energy as Equations

Work
W = F s cosθ
θ is the angle between the force and the direction of motion; when they are 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 work is found from the area under the graph

From concept to problem

Example 1: Work and Final Speed
On a frictionless surface, a 2 kg object at rest is pushed by a constant force of 4 N in the direction of motion over a distance of 4 m. Find the work done by the force and the object's final speed.
1
Find the work done by the constant force.
W = F s = 4 × 4 = 16
2
The object started at rest, so use the work–energy theorem to find the final speed.
16 = 12(2)v2 → v2 = 16 → v = 4
▸ work 16 J, final speed 4 m/s
All the work done by the net force becomes kinetic energy. With friction, some of it would escape as heat and the final speed would be lower.

Exam Points

CSAT-style problem (Physics II), adapted
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 is the area under the F-x graph. The rectangle (0 to 4 m) has area 10 × 4 = 40 J.
2
The triangle (4 to 6 m) has area ½ × 2 × 10 = 10 J, so the total is 50 J. So ① and ② are wrong, and ④ is wrong too, because a force that is present still does work.
3
By the work–energy theorem ΔEk = W = 50 J, and the change does not depend on the initial speed, so ⑤ is wrong and ③ is correct.
🎯 Exam Points
①Work W = Fs cosθ, and W = 0 when the force is perpendicular to the motion
②Work–energy theorem: work done by the net force = ΔE_k = ½mv² − ½mv₀²
③The work done by a changing force is the area under the F-x graph
④The change in kinetic energy is set by the work done by the net force alone, whatever the initial speed
⑤The work done by friction is negative, so it reduces the kinetic energy
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