Derive kinetic-energy change from Newtonian dynamics

Start with a declared system and frame

For a constant-mass point object in an inertial frame, Fnet=m dv/dt. Dot both sides with displacement dr=v dt: Fnet·dr=m v·dv. In fixed Cartesian axes, v·dv=vx dvx+vy dvy+vz dvz, whose integral is one-half the change of the squared speed. Thus net work equals the change of K=mv²/2. This derivation does not require a constant force.

Separate scalar energy from vector motion

Kinetic energy is nonnegative, but its change and work may be positive, negative or zero. A negative velocity component does not make energy negative. A force perpendicular to instantaneous velocity has zero instantaneous power and can change direction without changing speed. Conversely, energy alone supplies a speed magnitude, not the velocity direction or travel time.

Integrate a position-dependent interaction

The workbench stipulates a net force F(x)=A−bx, mass 2 kg and K(0)=10 J while the object moves rightward. Integration gives K(x)=10+Ax−bx²/2. At x=2 m, work is 2A−2b J and speed magnitude is sqrt(K) in the selected SI units. Since this concave energy function has its minimum over the finite interval at an endpoint, checking both endpoints establishes that every supported recipe remains accessible.

Keep frame and system energy distinct

Under a constant-velocity frame change, speed and K generally change. Work-energy bookkeeping must use the same frame for forces, displacement and kinetic energy. For several particles, total translational kinetic energy is the sum of their individual mv²/2 values; it is not generally just M times center-of-mass speed squared over two. Two equal masses moving oppositely can have a stationary center and positive total kinetic energy.

Explain before using the output

For A=4 N and b=1 N/m, work to 2 m is 6 J; K rises from 10 to 16 J and positive speed from sqrt(10) to 4 m/s. At A=2,b=2, the net force becomes negative after x=1, but the object retains positive K. Acceleration changes sign before the velocity necessarily does. A zero-work interval can also contain positive and negative contributions that cancel.

Worked original case

A=4,b=1 gives W=6 J,Kf=16 J,vf=4 m/s. A different mass would change the speed at the same energy, not the definition of work.

Investigation

Compare A=2,b=0 and A=2,b=2. Derive the complete K(x), locate any force zero, and explain why endpoint work does not describe every local change. Retain both captures.

Track force-point displacement when the system changes shape

First check a constant oblique force F=(3,4) N acting through displacement (2,0) m: its work is 6 J, not the force magnitude 5 N times distance 2 m. Only the force component along displacement contributes.

Now select an ideal symmetric spring system whose center remains fixed while both ends move outward by s from 0 to .1 m. In a reversible quasistatic virtual protocol with negligible final kinetic energy, each external force increases as F(s)=40s N, from 0 to 4 N; its average over the displacement is 2 N. The left force and left displacement are both negative, so their work is positive. Each endpoint contributes integral 0..0.1 40s ds=.2 J, for total input .4 J. A spring with k=20 N/m extends by2s=.2 m and stores k(.2)²/2=.4 J. The external forces have zero resultant and the center does not move, yet the internal configuration and energy change. Applying net force times center displacement would incorrectly return zero for this total work.

Predict the changed case s=.05 m: each endpoint integral is .05 J, total .1 J, matching the spring energy at extension .1 m. The assigned force history and quasistatic limit justify the storage claim; a suddenly applied constant force could leave internal kinetic energy and would require another account. This also illustrates why a system’s internal shape may change under external interactions even when a center-only motion model remains unchanged.

Public numerical transfer practice

Try a new practice case

A 2 kg point mass has velocity components (3,4) m/s. What is its kinetic energy in joules?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original teaching, values and calculations, checked 30 October 2026 against the official calculus-based Mechanics framework and OpenStax work and kinetic energy. The numerical packets are supplied or generated, not measured apparatus records. No physical launch, machinery, loading or field intervention is requested. Public feedback and virtual captures do not establish independent mastery or supervised laboratory participation.