Work and Kinetic Energy

Work and the work-energy theorem

Work measures energy transferred to or from a particle by a force while the point of application is displaced. For a constant force acting during a straight-line displacement, the definition is

W = F \cdot Δ r = F Δ r cos θ .

The dot product selects the component parallel to the displacement. If $\overset{s}{^} = Δ r /Δ r$ , then

W = (F \cdot \overset{s}{^}) Δ r = F_{∥} Δ r .

Work is a scalar; its SI unit is the joule, $1 J = 1 N m$ . Force remains a vector measured in newtons.

The sign follows from the geometry.

Tow rope pulling a sled forward: positive work.
Kinetic friction on a sliding crate: negative work.
Normal force on an object crossing a smooth horizontal surface: zero work, though it still enforces the constraint.
Pendulum tension: zero work, because the instantaneous displacement is tangential and the tension radial.

A constant force pulls a block at an angle across a level surface; only its horizontal component does work over the horizontal displacement, while the normal force and weight stay perpendicular to the motion.

The scalar-product form remains valid in three dimensions. For components,

F \cdot Δ r = F_{x} Δ x + F_{y} Δ y + F_{z} Δ z .

On an incline, each force contributes through its component along the displacement; the magnitudes alone do not fix the work. A vertical rise has $Δ x = 0$ but nonzero work by weight; horizontal travel has $Δ y = 0$ but nonzero work by a horizontal applied force. When a force changes direction along the path, the constant-force formula no longer applies.

Variable force and the work integral.

Over a sufficiently short displacement $d r$ , the force is approximately constant, giving

d W = F (r) \cdot d r .

Adding the segments and taking the limiting sum gives the line integral

W_{C} = C \int F \cdot d r .

The subscript $C$ matters. For a general force field, work can depend on the path as well as on the endpoints. Along one-dimensional motion on the $x$ axis this reduces to

W_{x_{i} \to x_{f}} = x_{i} \int x_{f} F_{x} (x) d x .

The signed area under an $F_{x}$ versus $x$ graph is therefore work. Areas above the $x$ axis are positive; areas below are negative. The graph need not describe a force proportional to position. A constant force produces a rectangle, a linear spring produces a triangle over an interval beginning at equilibrium, and arbitrary curves require integration or a numerical area estimate.

Work is the signed area under a force-position curve. The area from

x_{a}

x_{b}

is positive and the area from

x_{b}

x_{c}

is negative, so the two regions enter the work with opposite signs.

For measured force-position data, register force and position to the same physical configuration before estimating an area. A sensor offset creates a work error that grows with the displacement interval; position lag shifts a sharp force feature into the wrong interval.

A spring obeying Hooke's law, $F_{x} = - k x$ , does work from $x_{i}$ to $x_{f}$ of

W_{s} = x_{i} \int x_{f} (- k x) d x = - \frac{1}{2} k (x_{f}^{2} - x_{i}^{2}) .

The minus sign reflects the restoring direction. Stretching the spring from zero to positive $x$ requires an external force to do positive work; the spring does negative work over that displacement. Releasing the stretched spring reverses the sign of its work as it moves toward $x = 0$ .

Kinetic energy and the work-energy theorem.

Newton's second law gives a direct connection between net work and speed. For one-dimensional motion,

F_{net} = ma, a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} .

Multiplication by $d x$ and integration yields

x_{i} \int x_{f} F_{net} d x = v_{i} \int v_{f} m v d v = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} .

Thus

W_{net} = Δ K, K = \frac{1}{2} m v^{2} .

The theorem sums work from every force: applied forces, gravity, friction, normal forces, tension, and any other interaction. Begin with $W_{net}$ , then substitute the individual work contributions:

j \sum W_{j} = K_{f} - K_{i} .

The speed rather than the velocity appears in $K$ . Reversing a particle's direction at unchanged speed leaves its kinetic energy unchanged, even though its velocity changes. A net force perpendicular to the velocity changes direction but performs no instantaneous work. Uniform circular motion provides the standard example: centripetal force is perpendicular to $v$ , so $d W = 0$ and the speed is constant.

On a circular path the velocity is tangent to the circle while the centripetal force points to the centre. That force stays perpendicular to the displacement, so it does no work and the speed is unchanged.

No constant-acceleration assumption enters: the result holds whenever the total braking work is known, including position-dependent forces. Doubling the initial speed quadruples both the required work and the stopping distance.

Power and multiple forces

Power is the rate at which work is done:

P_{av} = \frac{Δ W}{Δ t}, P = \frac{d W}{d t} .

Using $d W = F \cdot d r$ gives the instantaneous mechanical-power relation

P = F \cdot v .

One watt is one joule per second. For a force parallel to velocity, $P = F v$ ; a motor that exerts the same tractive force requires more power at greater speed. For a force perpendicular to velocity, $P = 0$ . This is consistent with the work-energy theorem because

P_{net} = \frac{d K}{d t} .

Power is often quoted in kilowatts; the engineering horsepower conversion is $1 hp \approx 746 W$ . Efficiency compares output power with input power, but no efficiency factor enters a work-energy calculation unless the system boundary is stated.

Power measurements require force and velocity from the same instant and frame. A delayed velocity channel can flip the sign during a reversal. For rotating machinery the analogous pair is torque and angular velocity, with the same timing requirement.

Several forces and constraints.

A particle subject to several forces has additive work because the dot product and integral are linear:

W_{net} = C \int (j \sum F_{j}) \cdot d r = j \sum C \int F_{j} \cdot d r .

Compute net work either by adding forces before integration or by adding the work of each force afterward. The latter form is often clearer when gravity, friction, and an applied force have distinct physical roles. A sign convention is still required for every scalar component. Along an incline at angle $α$ , the work of gravity over a displacement $s$ upward along the surface is

W_{g} = (- m g sin α) s = - m g (s sin α) = - m g Δ y .

The normal force is perpendicular to the surface displacement and contributes zero work. In force analysis it still fixes the contact constraint and sets the kinetic-friction magnitude $f_{k} = μ_{k} N$ when sliding occurs. A constraint force can have nonzero work if the point of application has displacement along the force, as in a moving support or a deforming system.

The work-energy theorem also applies when a particle travels on a curved path. In a short interval, $d r = v d t$ , so

d W = F \cdot v d t, \frac{d K}{d t} = F_{net} \cdot v .

Tangential force changes speed; perpendicular force changes direction. Decomposing $F = F_{t} \hat{t} + F_{n} \overset{n}{^}$ gives

\frac{d K}{d t} = F_{t} v .

The normal component has no instantaneous power. In circular motion, a purely centripetal force can be large while its work is identically zero.

Substituting the initial value $12 N$ for the variable force gives the wrong answer; its average over the interval is $8 N$ , which the area calculation accounts for automatically.

Diagnostic checks.

Units: a force law integrated over distance must retain a factor of length and yield joules.
Sign: $W_{net} > 0$ raises $v^{2}$ , $W_{net} < 0$ lowers it. A force perpendicular to the motion contributes no work however large its magnitude.
Direction: the theorem predicts speed only; the direction of the final velocity needs a separate force or constraint analysis.
Path vs. displacement: net displacement suffices only when the force is truly constant in magnitude and direction. Kinetic friction follows the local motion and needs a path integral proportional to distance travelled.
System boundary: it fixes whether a force is an internal conversion or an external transfer; the numerical work is unchanged.

Work along changing paths

The line integral is needed whenever the force direction changes along a path, even at constant magnitude. A force tangent to the path does work equal to its magnitude times path length; a force radial from a fixed centre, acting on fixed-radius circular motion, stays perpendicular to every local displacement and does zero work. Each short segment contributes a scalar product, and the full work is their signed sum.

Whether tension or contact does work depends on the displacement of its point of application. A taut string does zero work on a mass on a fixed-radius arc because the tension is radial, but does work if its length changes, since the application point then moves partly along the tension. A normal force on a fixed smooth track does zero work; a moving support transfers energy through the same contact.

Power as a component measurement.

Instantaneous power is the force component along the current velocity times speed. A traction force parallel to motion gives positive power; a braking or drag force, negative power; a sideways constraint force, zero. The same force flips from positive to negative power if the motion reverses while the force direction holds.

Power is not energy. A large power over a short time can transfer less energy than a smaller power sustained longer. The area under a power-time graph is work, as is the area under a force-position graph; for recorded motion, work follows from integrating either force along the path or power over time.

The area under a power-time history is the work transferred. Positive power adds kinetic energy and negative power removes it, so the signed areas, not the peak power alone, set the net work over the interval shown.

Work-energy checks and stopping feasibility.

A brake, surface, or resisting field removes kinetic energy through the negative work available over the stated distance. If the initial kinetic energy exceeds that capacity, the object cannot stop within the region, and a signed energy balance shows this without a constant-acceleration assumption.

Write each work source separately before forming the net: an uphill displacement gives negative work by weight, kinetic friction gives negative work while sliding, a motor gives positive work. The final kinetic energy must stay nonnegative; a negative value signals that the presumed direction, interval, or stopping location is not reachable under the model.

Signed line-integral work on a path with reversals.

The differential displacement carries the direction of motion in a line integral. In one dimension, a path from $x = 0$ to $x = 4$ and back to $x = 1$ requires separate oriented segments. Forward and return segments have opposite $d x$ , so a fixed positive $x$ force component does positive work outward and negative work on return. With a constant force, net work depends on final displacement; signed segment contributions show transfer along the full path.

Parameterizing a path makes the orientation explicit. If a curve is described by $r (u)$ as parameter $u$ increases, then

W = u_{i} \int u_{f} F [r (u)] \cdot \frac{d r}{d u} d u .

Reversing the path reverses the sign of the integral for the same force field. A constant force does zero work on a closed path, since its net displacement is zero, even though the outward and return segments carry opposite signed works.

A constant force does positive work on the outward leg of a path and negative work on the return, because the local displacement reverses. Over the closed excursion shown the oriented contributions cancel and the net work is zero despite the nonzero path length.

Variable-force graphs: signed areas and interval boundaries.

An $F_{x}$ -- $x$ graph must be read as a definite integral. A force that changes sign at an interior point contributes positive area on one side and negative area on the other, and a region below the axis removes kinetic energy rather than being discarded.

Piecewise records integrate exactly with rectangles, triangles, and trapezoids; curved experimental data give a trapezoidal estimate whose precision matches the graph resolution. The horizontal axis must be displacement, not time: the area under a force-time graph is impulse, while the area under a force-position graph is work.

Under a position-only force law, reversing the direction of travel over the same coordinate reverses the sign of the work. A velocity-dependent force such as kinetic friction or drag is not a single-valued $F (x)$ curve through a forward-and-return cycle, so the graph must name the segment or carry the velocity dependence.

Treating the final segment as a positive area would give $41 J$ and overpredict the kinetic-energy increase. The negative force component matters even though the object continues in the positive coordinate direction.

Instantaneous power and conservative forces

The dot-product formula $P = F \cdot v$ is local: it uses force and velocity at the same instant and stays valid when either varies. Decomposing force into tangent and normal components gives $P = F_{t} v$ , so only the tangent component contributes. A cornering vehicle can carry a large sideways tire force that changes no kinetic energy while its forward component controls the speed.

A motor at maximum output power has available tractive force $F_{∥} = P / v$ , which falls as speed rises. At low speed the expression breaks down without a separate torque or traction limit, since real drives have finite force capacity. At high speed drag power grows quickly, leaving less net power for acceleration even at constant motor output.

The sign of power follows the force component along velocity. A braking force gives negative power and removes kinetic energy. A force perpendicular to velocity gives zero power even if it changes heading. The total instantaneous power of all forces equals the rate of kinetic-energy change, so a zero net power reading implies constant speed, not necessarily zero acceleration.

The force and velocity components set the instantaneous power. The forward part of the drive force adds power, the opposing drag removes power, and the perpendicular steering force changes heading while adding no power.

Work by gravity and by a spring.

Near Earth's surface, gravity is approximately a constant vector. Its work depends only on the vertical displacement between the endpoints, even when the actual path is curved or contains horizontal segments. With upward positive,

W_{g} = - m g (y_{f} - y_{i}) .

Moving upward gives negative work by gravity, moving downward positive; path length does not enter. Carrying a box up a long ramp and lifting it vertically through the same height give the same gravitational work under the constant- $g$ model, though other forces can do different work along the two routes.

A spring is a position-dependent force whose restoring force grows linearly with extension or compression from the unstretched reference. Its work follows from integrating the changing force, not from multiplying a final force by the full displacement: negative while the spring stretches away from equilibrium, positive while it returns.

Gravity and a linear spring are both position-dependent forces. Near the ground the gravity force is constant with height, so its force-height graph is flat, while the spring force grows in proportion to extension and its triangular area gives the work.

Each work comes from its own force law and displacement; they are not yet summed into a conservation statement.

Constraint-force work and moving contacts.

A constraint force does no work only when its point of application moves perpendicular to it. A normal force on a fixed smooth track meets this because the permitted displacement is tangent to the track; tension in a fixed-length pendulum string stays radial while the bob moves tangent to the arc. These zero-work results follow from geometry, not from the labels normal or tension.

When the constraint itself moves or deforms, it can do work. A moving conveyor's normal force does positive work on a package; a winch that changes a string's length does work through its tension. The work is evaluated from the actual motion of the application point in the chosen frame, so treating every constraint as zero-work would discard real transfers from motors, moving supports, and driven strings.

In a multi-body system a contact force can do negative work on one body and equal positive work on another. Compute each with its own point motion first; for an ideal internal constraint the two cancel in the combined-system balance. The force arrow alone means nothing without the receiving body, its displacement, and the system boundary.

A fixed smooth guide carries a normal force perpendicular to the particle's tangent displacement, so it does zero work. A moving support instead pushes along the object's displacement, transferring energy through the same kind of contact constraint.

Detailed work integrals

When a force law is given algebraically, work follows from integrating its component along the displacement coordinate over the stated limits. Units give a dimensional check: each term in $F_{x} (x)$ must have units of force, and integrating over position must produce joules. A constant term is a rectangle, a term proportional to $x$ a triangle, higher powers curved areas.

The limits carry the physical interval. Integrating from zero when the object begins at a nonzero coordinate adds work from a segment that never occurred. A negative integral does not mean a negative force magnitude; it means the force component opposed the oriented displacement on balance. Several variable forces can be summed separately or added first and integrated once.

Using a final force times the displacement would ignore the curvature of the force law. Using limits from zero to four metres would include an extra, untravelled segment. The integral states both the force model and the actual path interval.

Integral audit.

A variable-force integral needs the force-law coordinate, the actual endpoints, and the sign of the force component along the increasing coordinate; a graph shows the same audit, with positive area raising kinetic energy and negative area lowering it. When forces are given in component form, integrate only the component parallel to the differential displacement, and integrate each oriented segment in its actual direction so a return segment reverses the limits.

A measured force law gives an area estimate rather than an exact value. A dense set of force-position points supports a trapezoidal or fitted-curve integral; averaging the first and last force values is justified only when the graph is linear between them.

Frame dependence and force direction

Kinetic energy depends on the inertial frame because speed depends on the observer. A package resting on a train floor has zero kinetic energy in the train frame and nonzero kinetic energy in the ground frame when the train moves; both descriptions are valid. The theorem holds in each frame provided force, displacement, velocity, work, and kinetic energy are all evaluated in that same frame. Mixing frames is an accounting error.

The change in kinetic energy also transforms between frames. If a frame moves at constant velocity $V$ relative to another, then

v^{'} = v - V .

The kinetic-energy difference is not generally the same number in the two frames, because the squared speed changes. Work by a constant force can likewise differ: the force acts through the displacement measured in the selected frame. A force that does no work on an object stationary in one frame can do work on the same object in another frame where it moves.

Frame selection specifies the meaning of an energy statement. A stopping distance measured relative to a road is naturally analysed in the road frame; a package sliding inside a moving carriage can be analysed in the carriage frame if the carriage moves at constant velocity. State the chosen frame before assigning numerical kinetic energies or work values, especially when a system includes moving supports.

The same block and force are described from two inertial frames. In the ground frame the block moves through a nonzero displacement, while in the train frame it stays put, so the work and kinetic energy differ between frames even though the force arrow is identical.

Net work, speed change, and force direction.

Net work fixes the change in speed squared: positive net work raises the speed, negative net work lowers it, zero net work leaves it unchanged. The direction of the final velocity and the force responsible still need a vector analysis. A force component parallel to velocity changes speed; a perpendicular component changes heading only. A general force has both, curving the path while speeding the object up or slowing it down, and the net-work theorem sums their energy effects without reconstructing the trajectory.

A velocity reversal occurs at a turning point where the speed reaches zero. The work-energy equation locates it through $K = 0$ once the force work is known, but cannot select the direction afterward without the force past the turn. Uniform circular motion is the opposite case: nonzero net force, zero net work, since the force stays perpendicular to velocity.

Variable-force stopping distance.

Position-dependent braking stops the object when the accumulated negative work equals the initial kinetic energy. Write the signed resistive component along the direction of travel and integrate from the initial position to the unknown stopping position; the resulting equation may be linear, quadratic, or numerical, but the energy logic is unchanged.

Force variation can arise from a changing road grade, a spring buffer, a magnetic brake, or an energy-absorbing barrier. The sign follows the motion coordinate: for a vehicle moving in the positive direction a resistive force is negative, and a slope rising with distance adds a further negative gravitational component. An algebraic stopping root is physical only if it lies in the stated force-law interval and leaves nonnegative kinetic energy up to the stop.

A constant-force estimate using the initial $2000 N$ would predict $36 m$ , missing the increasing braking force by a wide margin.

Piecewise stopping models.

Real stopping regions often contain more than one force regime: a brake applies a constant force, then a spring buffer or gravel bed adds a position-dependent resistance. Work is computed in the physical order of travel. Integrate the force over the first segment, subtract that work from the initial kinetic energy, and carry any positive remainder into the next segment as its starting value. If kinetic energy reaches zero before a boundary, the later force law never engages. Each boundary is an event in the energy account; record the energy carried into it before selecting the next force law.

Event	Energy condition	Consequence
Entry to a new region	$K_{in} = K_{0} + \sum W_{prior}$	initialize the next integral
Stop before boundary	$K_{in} + W_{segment} = 0$	later region is not used
Exit with residual	$K_{out} > 0$	report speed and remaining path

A stop inside the first segment uses that segment's root, not a later formula; a root beyond the available barrier length leaves the object outside the region with residual kinetic energy, which the account states without assuming later motion. Every segment must share one inertial frame: a road-frame stopping distance uses the road-frame speed and displacement, while a moving walkway or vehicle floor changes both.

Stopping-model audit.

Before accepting a stopping root, check that its distance lies in the interval where the force law applies, that each work term has the correct sign, and that one inertial frame is used throughout. Audit a piecewise stop as an energy ledger indexed by location: the entry energy of each segment is the remaining kinetic energy from the previous one, not the initial energy copied into every force law.

Segment	Work evaluation	Physical check
Entry region	$W_{1} = \int_{x_{0}}^{x_{1}} F_{1} (x) d x$	root lies inside $[x_{0}, x_{1}]$
Later region	$W_{2} = \int_{x_{1}}^{x_{2}} F_{2} (x) d x$	use only if $K_{0} + W_{1} > 0$
Exit	$K_{f} = K_{0} + W_{1} + W_{2}$	report residual speed if $K_{f} > 0$

Reporting a work-energy result.

Report a work or kinetic-energy change with its system, reference frame, travel interval, force model, and sign convention. A number such as $- 18 kJ$ can mean work by a brake on a cart, work by the cart on the brake under the opposite convention, or one contribution inside a larger balance; the sentence must say which. Give force parameters and distances with units, and round only after the final balance is checked.

For measured force-position data, a constant sensor offset produces an area error proportional to path length, and a short unmeasured spike can dominate the work even when most points look smooth. Comparing against the measured change in kinetic energy gives an independent closure check when mass and speed are known. Any residual then belongs to one of:

a stated external work,
a changed potential energy,
measurement uncertainty,
a limitation of the chosen system boundary.