Work & the Work-Energy Theorem
- Define work as the line integral W = ∫F·dx and compute it for a variable force
- State and apply the work-energy theorem W_net = ΔKE
- Compute the work done by a position-dependent force such as a spring
Work is force integrated over displacement
For a constant force, work is just W = Fd cos θ — the force component along the motion times the distance. But most forces vary along the path (a spring pulls harder the more it is stretched). The general definition is a line integral: add up F·dx over every tiny step of the path. Graphically, W is the area under the force-versus-position curve, which is exactly what an integral computes.
The work-energy theorem
The net work done on an object — the work of the total force — equals the change in its kinetic energy, KE = ½mv². This single theorem replaces a whole chain of kinematics: if you know the net work, you know the change in speed, without ever finding the acceleration or the time. It follows directly from integrating ΣF = m dv/dt along the displacement.
A variable force F(x) = 3x² N acts on a 4 kg object as it moves from x = 0 to x = 2 m. If it starts from rest, find the work done and the final speed.
- 1.The force depends on position, so integrate: W = ∫₀² 3x² dx = [x³]₀² = 2³ − 0 = 8 J.
- 2.Apply the work-energy theorem with v_i = 0: W_net = ½mv_f² − 0.
- 3.Solve for the final speed: 8 = ½(4)v_f² = 2v_f², so v_f² = 4.
- 4.Therefore v_f = 2 m/s.
A variable force F(x) = 6x (N) acts on an object as it moves from x = 0 to x = 2 m. How much work does the force do?
The net work done on a 2 kg object initially at rest is 16 J. What is its final speed?
Whenever a force is given as a function of position, reach for the integral W = ∫F dx — never force times distance. The constant-force shortcut Fd cos θ is only a special case that fails the moment the force changes along the path.
Answer the 2 checkpoints as you read.
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