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Physics · Work, Energy, Momentum and Circular Motion

Kinetic Energy and the Work-Energy Theorem

Kinetic energy derived from Newton's second law, the work-energy theorem for constant and varying forces, its quadratic speed dependence, the link to momentum, and frictional dissipation.

Kinetic Energy and the Work–Energy Theorem

The previous chapter defined work as the transfer of energy by a force. This chapter identifies what that energy becomes when a body speeds up, and establishes the theorem that links the two: the net work done on a body equals the change in its kinetic energy. That single statement replaces, in many problems, the entire apparatus of forces and kinematics — and unlike the SUVAT equations, it remains exactly true when the force is not constant.

The theorem is also the first genuine bridge between the two halves of mechanics. It is derived from Newton's second law, so it contains no new physics; but it is expressed in scalars rather than vectors, and it compares two instants rather than tracking every moment between them. That change of viewpoint is what makes energy methods so much more economical than force methods for many questions.

The theorem. For any body, the net work done by all forces equals the change in kinetic energy:

Wnet=ΔEk=12mv2−12mu2W_{\text{net}} = \Delta E_k = \tfrac{1}{2}mv^{2} - \tfrac{1}{2}mu^{2}

1. Deriving Kinetic Energy

Consider a body of mass mm acted on by a constant resultant force FF along its direction of motion, which accelerates it from speed uu to speed vv over a displacement ss. Two results from Block A are all that is needed.

From Newton's second law, F=maF = ma. From the time-free kinematic equation, v2=u2+2asv^{2} = u^{2}+2as, so

a=v2−u22sa = \frac{v^{2}-u^{2}}{2s}

Substituting into the definition of work,

W=Fs=mas=m(v2−u22s)s=12mv2−12mu2W = Fs = mas = m\left(\frac{v^{2}-u^{2}}{2s}\right)s = \tfrac{1}{2}mv^{2} - \tfrac{1}{2}mu^{2}

The work done has produced a change in a quantity that depends only on the mass and the square of the speed. That quantity is the kinetic energy:

Ek=12mv2[J]E_k = \tfrac{1}{2}mv^{2} \qquad [\text{J}]

the energy a body possesses by virtue of its motion. It is a scalar, always positive or zero, and independent of the direction of travel — two cars of equal mass approaching each other at 20 ms20~\dfrac{\text{m}}{\text{s}} have identical kinetic energies despite opposite velocities.

The derivation above assumed a constant force, but the theorem survives without that restriction. Using the general definition of work and the chain-rule trick from the kinematics chapter, a=vdvdsa = v\dfrac{dv}{ds}:

W=∫F ds=∫m vdvds ds=∫uvmv dv=[12mv2]uvW = \int F\,ds = \int m\,v\frac{dv}{ds}\,ds = \int_u^v mv\,dv = \left[\tfrac{1}{2}mv^{2}\right]_u^v

The work–energy theorem holds for any force, constant or varying — which is precisely why it succeeds where the SUVAT equations fail.

2. Reading the Theorem

Three features of Wnet=ΔEkW_{\text{net}} = \Delta E_k deserve emphasis.

It is the net work. Every force acting must be included, with its sign. A crate dragged at constant velocity has positive work from the rope and equal negative work from friction; the net work is zero, and indeed ΔEk=0\Delta E_k = 0.

The sign tells you what happens. Positive net work speeds a body up; negative net work slows it down; zero net work leaves the speed unchanged (though possibly not the direction).

Only the endpoints matter. The theorem compares the initial and final states and is silent about everything in between. This is its great economy — and also its limitation, since it yields speeds but not times.

The quadratic dependence on speed has consequences worth internalising. Doubling a car's speed quadruples its kinetic energy, and therefore quadruples the work the brakes must do — which is the energy-based explanation of the result found kinematically in Block A, that braking distance goes as v2v^{2}. It is also why the difference between a 5050 and a 70 kmh70~\dfrac{\text{km}}{\text{h}} impact is far greater than the speed difference suggests.

3. Worked Examples

Example 1 — braking distance without kinematics. A car of mass 1250 kg1250~\text{kg} travelling at 24.0 ms24.0~\dfrac{\text{m}}{\text{s}} brakes to rest. The friction force is 7.20×103 N7.20\times 10^{3}~\text{N}. How far does it travel?

The only force doing work is friction, at 180∘180^\circ to the motion:

Wnet=−fs=ΔEk=0−12(1250)(24.0)2=−3.60×105 JW_{\text{net}} = -fs = \Delta E_k = 0 - \tfrac{1}{2}(1250)(24.0)^{2} = -3.60\times 10^{5}~\text{J} s=3.60×1057.20×103=50.0 ms = \frac{3.60\times 10^{5}}{7.20\times 10^{3}} = 50.0~\text{m}

No acceleration was computed and no kinematic equation was used. Note also that the mass cancels if the friction is written as μmg\mu mg: s=v22μgs = \dfrac{v^{2}}{2\mu g}, recovering the Block A result.

Example 2 — a varying force. A block of mass 2.0 kg2.0~\text{kg}, initially at rest, is pushed by a force that rises linearly from 00 to 30 N30~\text{N} over 4.0 m4.0~\text{m} on a frictionless surface. Find its final speed.

The work is the triangular area under the force–displacement graph:

W=12(4.0)(30)=60 JW = \tfrac{1}{2}(4.0)(30) = 60~\text{J} 12mv2=60⟹v=2(60)2.0=7.7 ms\tfrac{1}{2}mv^{2} = 60 \quad\Longrightarrow\quad v = \sqrt{\frac{2(60)}{2.0}} = 7.7~\frac{\text{m}}{\text{s}}

No SUVAT equation could have handled this, because the acceleration was not constant. This is the case where the theorem is not merely convenient but necessary.

Example 3 — a slope with friction. A sledge of mass 18 kg18~\text{kg} slides from rest 25 m25~\text{m} down a 12∘12^\circ slope with μk=0.11\mu_k = 0.11. Find its final speed.

Work by gravity, using the vertical drop h=25sin⁡12∘=5.20 mh = 25\sin 12^\circ = 5.20~\text{m}:

Wg=mgh=18(9.81)(5.20)=918 JW_g = mgh = 18(9.81)(5.20) = 918~\text{J}

Work by friction, with N=mgcos⁡12∘=173 NN = mg\cos 12^\circ = 173~\text{N}:

Wf=−μkNd=−0.11(173)(25)=−476 JW_f = -\mu_k N d = -0.11(173)(25) = -476~\text{J}

Net work, and hence the speed:

Wnet=918−476=442 Jv=2(442)18=7.0 msW_{\text{net}} = 918 - 476 = 442~\text{J} \qquad v = \sqrt{\frac{2(442)}{18}} = 7.0~\frac{\text{m}}{\text{s}}

4. Kinetic Energy and Momentum

Two quantities describe a body's motion, and confusing them is a classic error:

Ek=12mv2(scalar, J)p⃗=mv⃗(vector, kg⋅ms)E_k = \tfrac{1}{2}mv^{2} \quad (\text{scalar, } \text{J}) \qquad\qquad \vec{p} = m\vec{v} \quad (\text{vector, } \text{kg}\cdot\tfrac{\text{m}}{\text{s}})

They are related by a formula worth knowing, obtained by substituting v=pmv = \dfrac{p}{m}:

Ek=p22m⟺p=2mEkE_k = \frac{p^{2}}{2m} \qquad\Longleftrightarrow\qquad p = \sqrt{2mE_k}

This appears repeatedly — in collisions, in the photoelectric effect, and in the de Broglie wavelength of a particle accelerated through a potential difference.

The two behave very differently. Kinetic energy has no direction and cannot cancel, so two bodies moving in opposite directions have zero total momentum but a non-zero total kinetic energy. And in a collision, momentum is always conserved while kinetic energy generally is not — the distinction that classifies collisions as elastic or inelastic in a later chapter.

5. Where the Energy Goes: Dissipation

If negative net work removes kinetic energy, where does that energy go? It is not destroyed — energy never is. For friction, it appears as thermal energy in the two surfaces and the surrounding air:

ΔEthermal=f d\Delta E_{\text{thermal}} = f\,d

where dd is the distance travelled, not the displacement. This distinction matters: a block sliding back and forth over 10 m10~\text{m} of floor and returning to its start has zero displacement but has generated friction heating over the whole 10 m10~\text{m}.

This is precisely why friction is non-conservative and why no "friction potential energy" exists: the energy has been dispersed into the random motion of countless molecules and cannot be recovered as organised kinetic energy. The general accounting statement,

Wnet, external=ΔEk+ΔEthermal+ΔEp+…W_{\text{net, external}} = \Delta E_k + \Delta E_{\text{thermal}} + \Delta E_p + \ldots

is the conservation of energy, taken up properly two chapters ahead.

6. When to Use Energy and When to Use Forces

Energy methods and force methods answer different questions, and choosing correctly saves a great deal of time.

Use the work–energy theorem when:

  • the question asks for a speed or a distance and does not mention time;
  • the force varies with position (springs, non-uniform fields, measured curves);
  • the path is complicated but only the endpoints matter (a bead on a curved wire, a rollercoaster).

Use Newton's laws when:

  • the question asks for a time, an acceleration, or a force at a particular instant;
  • you need the normal force or the tension at some point;
  • the motion involves directions that energy, being a scalar, cannot distinguish.

Many problems need both: an energy argument to find the speed at a point, then a force argument to find the normal force there. That combination is the standard structure of a rollercoaster or vertical-circle question in Block B.

7. Common Exam Pitfalls

  • Using only one force's work instead of the net work. Every force acting contributes, with its sign.
  • Forgetting that Ek∝v2E_k \propto v^{2}. Doubling the speed quadruples the energy, and the braking distance with it.
  • Treating kinetic energy as a vector. It is a positive scalar; bodies moving in opposite directions have kinetic energies that add, not cancel.
  • Confusing EkE_k with momentum. Use Ek=p22mE_k = \dfrac{p^{2}}{2m} to move between them rather than guessing.
  • Using the displacement rather than the distance for friction heating. The thermal energy generated is f df\,d along the actual path.
  • Applying SUVAT to a varying force. The work–energy theorem still works there; the kinematic equations do not.
  • Assuming zero net work means no forces act. It means they cancel — a body at constant velocity has several forces and zero net work.
  • Expecting the theorem to give a time. It relates work to speed change and contains no time at all.

Summary

  • Kinetic energy Ek=12mv2E_k = \tfrac{1}{2}mv^{2} is the energy of motion: a positive scalar in joules, independent of direction.
  • The work–energy theorem, Wnet=ΔEkW_{\text{net}} = \Delta E_k, follows from F=maF = ma combined with v2=u2+2asv^{2} = u^{2}+2as, and holds for any force, constant or varying.
  • It is the net work that matters, including every force with its sign; positive net work speeds a body up, negative slows it down.
  • Because Ek∝v2E_k \propto v^{2}, doubling the speed quadruples the energy — the energy explanation of why braking distance goes as v2v^{2}.
  • Kinetic energy and momentum are linked by Ek=p22mE_k = \dfrac{p^{2}}{2m}, but differ fundamentally: one is a scalar that cannot cancel, the other a vector that can.
  • Energy removed by friction becomes thermal energy f df\,d, computed along the distance travelled, and cannot be recovered — the reason friction is non-conservative.
  • Use energy methods for speeds, distances and varying forces; use Newton's laws for times, accelerations and instantaneous forces. Many problems require both.