Work-Energy Theorem

Study a point mass initially moving along +x on a frictionless line under constant net force F.

Work-Energy Theorem: connecting force, displacement, and kinetic energy

The work-energy theorem states that the net work done on an object equals its change in kinetic energy. This principle allows us to determine the final speed of an object after a constant force has been applied over a specific distance without having to use kinematic equations directly.

Net Work and Kinetic Energy

For a constant net force F applied over a displacement x, the work done is W = F · x.

The kinetic energy of an object with mass m and speed v is K = ½mv². The theorem connects these: W = K - K₀, where K₀ is the initial kinetic energy.

Worked example: K₀=9J, work=8J, K=17J

Consider a point mass of 2 kg initially moving at 3 m/s. A constant net force of 4 N acts in the direction of motion for 2 m.

Calculate initial kinetic energy: K₀ = ½(2)(3)² = 9 J.
Calculate work done: W = F · x = 4 · 2 = 8 J.
Find final kinetic energy: K = K₀ + W = 9 + 8 = 17 J.
Find final speed: v = √(2K/m) = √(34/2) = √17 ≈ 4.123105 m/s.

If instead the force is -3 N over 4 m with initial speed 4 m/s, work is -12 J and final K is 16 - 12 = 4 J, resulting in v = 2 m/s.

Experiment with the Work-Energy Module

Open Work-Energy Theorem. The controls provided are Mass m, Initial speed v0, Net force F, and Requested displacement x.

Set Mass m to 2, Initial speed v0 to 3, Net force F to 4, and Requested displacement x to 2. Observe the calculated work of 8 J and final speed of √17.
Set Mass m = 2, Initial speed v0 = 4, Net force F = -3 and Requested displacement x = 4. Initial energy is 16 J, requested work -12 J, K at request 4 J and speed 2 m/s.
Set Mass m = 1, Initial speed v0 = 0, Net force F = 5 and Requested displacement x = 2. Work and final K are 10 J, and speed is √20, approximately 4.472135955 m/s.
Set Mass m = 2, Initial speed v0 = 2, Net force F = -2 and Requested displacement x = 3. The forward request is unreachable and the stopping point is 2 m. No final K or speed at x = 3 is shown.
Change Requested displacement x to 2. The stop is reachable with K = 0 and speed 0.
Set Mass m = 1, Initial speed v0 = 0, Net force F = 0 and Requested displacement x = 1. The object is stationary at the origin; no speed or K at the unvisited request is claimed. Set x = 0 to recover the valid initial state.
Clear Mass m and observe the accessible error with calculations removed. Restore m = 2, then choose Reset view after panning; inputs remain unchanged.

Limitations of the Model

This module simplifies the scenario to a single constant force acting on a point mass in one dimension on a frictionless line. It does not account for variable forces (where work requires integration), two- or three-dimensional motion, or rotational kinetic energy. The calculations also focus on the initial forward branch; motion after a reversal at the stopping point is not modeled. The stationary case (zero initial speed and zero net force) is considered unreachable for any positive displacement, while x = 0 remains the initial state. At an unreachable point the product F × x is a formal requested work value, not work realized by reaching that point. A stopping point beyond the 0–10 m diagram is reported numerically rather than moved onto the diagram. Extremely small nonzero forces can place it beyond representable numeric range. No negative energy or speed is silently clamped into a physical state.

References

For a detailed theoretical derivation and further examples, refer to OpenStax University Physics Volume 1, Section 7.3: Work-Energy Theorem.