Work & Kinetic Energy

Energy is one of the central concepts of physics, and kinetic energy $K$ is the most direct form — it is the energy an object possesses by virtue of its motion. A speeding locomotive, a rotating turbine blade, a thrown baseball: each can do work on other objects because of its motion. Kinetic energy measures exactly how much.

Here $m$ is the object's mass in kilograms and $v$ is its speed in m/s. Two essential properties: (1) $K$ is a scalar — it has no direction. (2) $K \geq 0$ always — a stationary object has $K = 0$ and there is no such thing as negative kinetic energy.

The SI unit of kinetic energy — and all forms of energy — is the joule (symbol J), named after the 19th-century English physicist James Prescott Joule. From the formula, $1\,\text{J} = 1\,\text{kg}\cdot\text{m}^2/\text{s}^2$. To put this in perspective: a 3 kg duck flying at 2 m/s has $K = \frac{1}{2}(3)(4) = 6\,\text{J}$; a 70 kg sprinter running at 10 m/s has $K = \frac{1}{2}(70)(100) = 3500\,\text{J}$.

Defining Work

When a force acts on an object and the object moves, the force may transfer energy to or from the object. This transfer of energy by a force is called work. Energy transferred to the object is positive work; energy transferred from the object is negative work. Work, like energy, is a scalar measured in joules.

For a constant force $\vec{F}$ acting on an object that undergoes displacement $\vec{d}$, the work done by the force is:

where $\phi$ is the angle between the directions of the force $\vec{F}$ and the displacement $\vec{d}$. The second form uses the dot product, which is especially convenient when vectors are given in unit-vector notation: $\vec{F} = F_x\hat{i} + F_y\hat{j}$ and $\vec{d} = d_x\hat{i} + d_y\hat{j}$ gives $W = F_x d_x + F_y d_y$.

Sign of Work

The sign of work follows directly from $\cos\phi$:

• $\phi < 90°$: $\cos\phi > 0$, so $W > 0$ — force has a component in the direction of motion. Energy is transferred to the object (it tends to speed up).
• $\phi = 90°$: $\cos 90° = 0$, so $W = 0$ — force is perpendicular to motion. No energy transfer.
• $\phi > 90°$ (up to 180°): $\cos\phi < 0$, so $W < 0$ — force opposes motion. Energy is transferred from the object (it tends to slow down).

Net Work

When multiple forces act on an object, the net work $W_\text{net}$ is the sum of the works done by all individual forces. Equivalently, it is the work done by the net force $\vec{F}_\text{net}$. Both methods give the same result.

The Work–Kinetic Energy Theorem

The most powerful result of this chapter connects work to the change in kinetic energy. For any particle (an object treated as a point mass):

Or rearranged: $K_f = K_i + W_\text{net}$. In words: the kinetic energy after all work is done equals the kinetic energy before, plus the net work done. If $W_\text{net} > 0$, the object speeds up. If $W_\text{net} < 0$, it slows down. If $W_\text{net} = 0$, the speed is unchanged.

This theorem is a scalar equation — no directions, no components. When you care about speed at two points (not the full trajectory), it is often faster than applying Newton's second law. The theorem holds for constant or variable forces and for any path, straight or curved.

Gravity is one of the most commonly encountered forces, so its work deserves special attention. For a particle-like object of mass $m$ moving through displacement $\vec{d}$, the work done by the gravitational force $\vec{F}_g = -mg\hat{j}$ (pointing downward) is:

where $\phi$ is the angle between the gravitational force direction (downward) and the displacement $\vec{d}$.

Rising and Falling Objects

For an object moving upward through distance $d$: the displacement is upward but gravity is downward, so $\phi = 180°$ and $\cos 180° = -1$:

The minus sign confirms that gravity removes kinetic energy from a rising object — which is why thrown balls slow down. For an object falling through distance $d$: both gravity and displacement point downward, $\phi = 0°$:

Gravity adds kinetic energy to a falling object — which is why falling objects speed up.

Lifting and Lowering an Object

When you lift an object with an applied force $\vec{F}_a$, both the applied force and gravity act on it. Applying the work–kinetic energy theorem:

If the object starts and ends at rest (or at the same speed), then $\Delta K = 0$, and the equation reduces to:

This says that when you slowly lift an object from the floor to a shelf — even if you vary the force during the lift — the total work your hands do equals $mgd$. You do not need to know how the force varied, only the endpoints. This is one of the great simplifications that energy methods provide.

The spring is the prototypical variable force — one that changes in magnitude as the object moves. Many forces in nature (molecular bonds, elastic materials, suspension systems) behave like springs over some range, so mastering this one case unlocks understanding of many others.

Hooke's Law

For a spring with one end fixed, if we define $x$ as the displacement of the free end from its relaxed position (neither compressed nor extended), then the spring force on an object attached to the free end is:

The constant $k$ is the spring constant (or force constant), measured in N/m. A large $k$ means a stiff spring that exerts large forces for small displacements. The minus sign is critical: the force always opposes the displacement. Pull the spring right ($x > 0$) and it pulls back left ($F_s < 0$); push it left ($x < 0$) and it pushes back right ($F_s > 0$). This is why it is called a restoring force — it always acts to restore the spring to $x = 0$.

Work Done by the Spring Force

Because $F_s = -kx$ varies with position, we cannot use $W = Fdcosphi$ directly — there is no single value of $F$. Instead we integrate:

Three cases to remember: (1) if the object ends closer to $x = 0$ than it started ($|x_f| < |x_i|$), then $W_s > 0$ (spring does positive work, object speeds up). (2) If the object ends farther from $x = 0$, then $W_s < 0$. (3) If $x_i = 0$ and the spring is stretched or compressed by $x$, the work done by the spring is $W_s = -\frac{1}{2}kx^2$ (always negative — the spring opposes the displacement).

If an object attached to a spring is stationary both before and after a displacement (e.g., you slowly stretch a spring), then $\Delta K = 0$ and the work–kinetic energy theorem gives $W_a + W_s = 0$, so $W_a = -W_s = \frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2$. You do the negative of the spring's work.

The spring is one example of a variable force. In general, many forces depend on position: gravity varies with altitude, electrostatic force varies with distance, air resistance varies with speed. How do we find the work done by an arbitrary $F(x)$ as an object moves from $x_i$ to $x_f$?

Integration as a Sum of Tiny Work Increments

The strategy is elegant: divide the displacement into tiny segments $\Delta x$ so small that $F(x)$ is nearly constant within each segment. Then the work done in segment $j$ is approximately $\Delta W_j = F_j\,\Delta x$ — just the familiar constant-force formula. Summing all segments:

Taking the limit as $\Delta x \to 0$ turns the sum into an integral:

Connecting Back to the Work–Kinetic Energy Theorem

Even with a variable force, the work–kinetic energy theorem still holds: $\Delta K = W_\text{net}$ where $W_\text{net}$ is found by integrating the net force over the displacement. This can be proven rigorously using Newton's second law and the chain rule of calculus, but the result is reassuringly simple: the theorem works whether forces are constant or not.

For motion in three dimensions, the work generalizes to a line integral: $W = \int \vec{F}\cdot d\vec{r}$, which separates into three one-dimensional integrals when force components depend only on their respective coordinates.

Two workers can do the same amount of work — say, carry 500 bricks up a flight of stairs — but one does it in 10 minutes and the other in 1 hour. Both do the same work, but the fast worker delivers more power. Power is the rate at which work is done, i.e., the rate at which energy is transferred.

Average and Instantaneous Power

If a force does work $W$ in time interval $\Delta t$, the average power is:

The instantaneous power is the time derivative of work:

For a constant force $\vec{F}$ acting on an object moving at instantaneous velocity $\vec{v}$, the instantaneous power has an elegant form. Since $dW = \vec{F}\cdot d\vec{r}$, dividing by $dt$ gives:

where $\phi$ is the angle between $\vec{F}$ and $\vec{v}$. This formula is especially useful when force and velocity are known at an instant (e.g., the power output of a car engine at a given speed and throttle).

Units of Power

The SI unit of power is the watt (W), named after James Watt, whose improvements to the steam engine powered the Industrial Revolution:

The formula $P = Fv$ reveals an important trade-off for machines: at a given power output, a larger force means a smaller velocity, and vice versa. A car engine at fixed power output must apply a smaller force to go faster — this is why your car accelerates quickly from rest but struggles to gain more speed at highway velocity. Bicycles exploit gear ratios to keep pedaling force and cadence (speed) both in a comfortable range as terrain changes.