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Physics · Ch 6 — Work, Energy and Power

Notions of Work and Kinetic Energy: The Work-Energy Theorem

6.2

Notions of Work and Kinetic Energy: The Work-Energy Theorem

Opening the Idea: Why Work and Kinetic Energy Belong Together

The chapter begins with the everyday notion of work — the harder you push and the farther something moves, the more work you do. But physics demands a precise, quantitative definition. The key insight is that work, defined as force times displacement in the direction of the force, is not an end in itself. It is a transfer of energy. When you do work on an object, you change its state of motion — specifically, you change its speed. That change in speed is tied to a quantity called kinetic energy.

The work-energy theorem is the bridge: it states that the net work done on a particle equals the change in its kinetic energy. This is not a new law; it is a direct consequence of Newton's second law, derived by integrating the equation of motion along the path. It gives us a powerful alternative to using F=maF = ma directly, especially when forces vary or when we care only about speeds, not times.


Deriving the Work-Energy Theorem for a Constant Force

Start with the simplest case: a constant net force F⃗\vec{F} acts on a particle of mass mm along a straight line. The particle moves a displacement d⃗\vec{d}.

From Newton's second law, the acceleration is constant: a=F/ma = F/m (taking the direction of force as positive). The equations of motion for constant acceleration give the relation between initial velocity uu, final velocity vv, acceleration aa, and displacement ss:

v2=u2+2asv^2 = u^2 + 2as

Substitute a=F/ma = F/m and s=ds = d:

v2=u2+2(Fm)dv^2 = u^2 + 2\left(\frac{F}{m}\right)d

Multiply both sides by m/2m/2:

12mv2=12mu2+Fd\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + Fd

The term FdFd is the work done by the net force (since force and displacement are in the same direction, W=Fdcos⁡0∘=FdW = Fd \cos 0^\circ = Fd). Rearranging:

Fd=12mv2−12mu2Fd = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

This is the work-energy theorem for a constant force along a straight line. The quantity 12mv2\frac{1}{2}mv^2 is defined as the kinetic energy KK of the particle. So the theorem reads:

W=Kf−Ki=ΔKW = K_f - K_i = \Delta K

The net work done on a particle equals the change in its kinetic energy.

Note

This derivation assumes the force is constant and the motion is along a straight line. But the theorem is far more general, as we will see.


Defining Kinetic Energy

From the derivation, kinetic energy emerges naturally:

K=12mv2K = \frac{1}{2}mv^2

It is a scalar quantity, always positive or zero. It depends only on the speed of the particle, not on its direction. The SI unit is the joule (J), the same as work — 1 J=1 kg m2/s21\,\text{J} = 1\,\text{kg m}^2/\text{s}^2.

Watch out

Kinetic energy is not a vector. Even though velocity is a vector, squaring it removes directional information. A particle moving at 10 m/s to the left has the same kinetic energy as one moving at 10 m/s to the right.


The General Case: Variable Force Along a Curved Path

The constant-force, straight-line case is too restrictive. Real forces often vary with position (like a spring) or act along curved paths. The work-energy theorem holds in full generality, and the proof uses calculus.

Consider a particle moving along a curved path under the action of a net force F⃗\vec{F} that may vary from point to point. The work done by the net force as the particle moves from an initial position ii to a final position ff is:

W=∫ifF⃗⋅dr⃗W = \int_i^f \vec{F} \cdot d\vec{r}

where dr⃗d\vec{r} is an infinitesimal displacement along the path.

From Newton's second law, F⃗=ma⃗=mdv⃗dt\vec{F} = m\vec{a} = m\frac{d\vec{v}}{dt}. Also, dr⃗=v⃗ dtd\vec{r} = \vec{v}\,dt. So:

F⃗⋅dr⃗=mdv⃗dt⋅v⃗ dt=m v⃗⋅dv⃗\vec{F} \cdot d\vec{r} = m\frac{d\vec{v}}{dt} \cdot \vec{v}\,dt = m\,\vec{v} \cdot d\vec{v}

Now, v⃗⋅dv⃗=12d(v⃗⋅v⃗)=12d(v2)\vec{v} \cdot d\vec{v} = \frac{1}{2}d(\vec{v} \cdot \vec{v}) = \frac{1}{2}d(v^2). This is a key calculus identity: the differential of v2v^2 is 2v⃗⋅dv⃗2\vec{v} \cdot d\vec{v}.

Therefore:

F⃗⋅dr⃗=12m d(v2)\vec{F} \cdot d\vec{r} = \frac{1}{2}m\,d(v^2)

Integrate from initial state ii to final state ff:

W=∫ifF⃗⋅dr⃗=∫vivf12m d(v2)=12mvf2−12mvi2W = \int_i^f \vec{F} \cdot d\vec{r} = \int_{v_i}^{v_f} \frac{1}{2}m\,d(v^2) = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

This is the work-energy theorem in its most general form. It holds for any net force, any path, and any variation of force along the path.

Important

The theorem is a scalar relation. It connects the total work (a scalar sum of dot products) to the change in kinetic energy (a scalar). It does not give directional information about the final velocity, only its magnitude.


Properties and Consequences of the Work-Energy Theorem

The textbook lists several important properties that follow from the theorem. Each is derived or explained below.

Property (I): Work Done by a Force Can Be Positive, Negative, or Zero

The work done by a single force F⃗\vec{F} over a displacement dr⃗d\vec{r} is dW=F⃗⋅dr⃗=F drcos⁡θdW = \vec{F} \cdot d\vec{r} = F\,dr\cos\theta, where θ\theta is the angle between the force and the displacement.

  • If 0∘≤θ<90∘0^\circ \leq \theta < 90^\circ, cos⁡θ>0\cos\theta > 0, work is positive. The force has a component in the direction of motion, increasing speed and kinetic energy.
  • If θ=90∘\theta = 90^\circ, cos⁡θ=0\cos\theta = 0, work is zero. The force is perpendicular to motion (e.g., centripetal force), does no work, and does not change kinetic energy.
  • If 90∘<θ≤180∘90^\circ < \theta \leq 180^\circ, cos⁡θ<0\cos\theta < 0, work is negative. The force opposes motion, decreasing speed and kinetic energy.
Tip

When a force does negative work, it is often said that the object does work against that force. For example, when you lift a book, gravity does negative work on the book (the book's kinetic energy decreases if it slows down), while you do positive work on it.

Property (II): The Net Work is the Sum of Works Done by Individual Forces

If multiple forces act on a particle, the net work WnetW_{\text{net}} is the sum of the works done by each force:

Wnet=W1+W2+W3+…W_{\text{net}} = W_1 + W_2 + W_3 + \dots

This follows because the net force is the vector sum of individual forces, and the work integral is linear:

Wnet=∫F⃗net⋅dr⃗=∫(F⃗1+F⃗2+… )⋅dr⃗=∫F⃗1⋅dr⃗+∫F⃗2⋅dr⃗+…W_{\text{net}} = \int \vec{F}_{\text{net}} \cdot d\vec{r} = \int (\vec{F}_1 + \vec{F}_2 + \dots) \cdot d\vec{r} = \int \vec{F}_1 \cdot d\vec{r} + \int \vec{F}_2 \cdot d\vec{r} + \dots

The work-energy theorem then states:

Wnet=ΔKW_{\text{net}} = \Delta K

So the total change in kinetic energy is the sum of the works done by all forces acting on the particle. …