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Physics · Work, Energy, Momentum and Circular Motion

Definition of Mechanical Work

Work as the transfer of energy by a force: the cosine rule and its sign, the cases that give zero work, the area under a force-displacement graph, and conservative versus non-conservative forces.

Definition of Mechanical Work

Block A described motion and then explained it with forces. This block introduces a second and, in many ways, more powerful way of analysing the same situations: through energy. The advantage is immediate. Newton's laws require you to know the forces at every instant and to integrate their effect through time; an energy argument compares only the beginning and the end, and often gives in one line what a force analysis would take a page to produce.

Everything in the block rests on one definition, made in this chapter. Work is the mechanism by which energy is transferred from one system to another by a force. Getting its definition exactly right — including the angle, including the sign, including the cases where it is zero — determines whether every energy argument that follows is correct.

The definition. The work done by a constant force is the product of the magnitude of the force, the magnitude of the displacement, and the cosine of the angle between them:

W=Fscos⁡θ=F⃗⋅s⃗[J]W = Fs\cos\theta = \vec{F}\cdot\vec{s} \qquad [\text{J}]

1. Why the Cosine Appears

A force pushing a crate at an angle does not use all of its magnitude to move the crate along the floor. Only the component of the force parallel to the displacement contributes to the motion; the perpendicular component pushes the crate against the floor (or lifts it) and moves it nowhere.

Resolving the force into components parallel and perpendicular to the displacement,

F∥=Fcos⁡θF⊥=Fsin⁡θF_\parallel = F\cos\theta \qquad F_\perp = F\sin\theta

and the work done is the parallel component multiplied by the displacement:

W=F∥ s=Fscos⁡θW = F_\parallel\,s = Fs\cos\theta

The identical result may be read the other way round — as the force multiplied by the component of the displacement in the direction of the force — which is sometimes the more natural reading, for instance when computing the work done by gravity on a body sliding down a slope.

This is the scalar product of the force and displacement vectors, and it is worth noting that the product of two vectors here yields a scalar: work has magnitude but no direction. It does, however, have a sign, and that sign carries physical meaning.

2. The Joule, and the Sign of Work

The SI unit of work, and therefore of energy, is the joule:

1 J=1 N⋅m=1 kg⋅m2s21~\text{J} = 1~\text{N}\cdot\text{m} = 1~\frac{\text{kg}\cdot\text{m}^{2}}{\text{s}^{2}}

One joule is the work done when a force of one newton moves its point of application one metre in the direction of the force.

The sign of cos⁡θ\cos\theta determines the sign of the work, and the three cases are physically distinct:

  • θ<90∘\theta < 90^\circ gives W>0W > 0. The force has a component along the motion and transfers energy to the body. An engine driving a car forward does positive work.
  • θ=90∘\theta = 90^\circ gives W=0W = 0. The force is perpendicular to the motion and transfers no energy at all.
  • θ>90∘\theta > 90^\circ gives W<0W < 0. The force opposes the motion and removes energy from the body. Friction and air resistance always do negative work on a moving body.

Negative work is not a bookkeeping oddity; it is the statement that energy is flowing the other way. When a crate slides to a halt, the friction has done negative work equal in magnitude to the kinetic energy the crate lost, and that energy has appeared as thermal energy in the crate and the floor.

3. When No Work Is Done

Three distinct situations give zero work, and distinguishing them is regularly examined.

No displacement. A person straining to hold a heavy suitcase stationary does no work on the suitcase, however tired they become. Their muscles consume chemical energy in maintaining tension, but no energy is transferred to the suitcase because it does not move. Work requires displacement, not effort.

Force perpendicular to displacement. Carrying that suitcase horizontally at constant velocity, the upward force does no work, since cos⁡90∘=0\cos 90^\circ = 0. Similarly:

  • The normal force never does work on a body sliding along a surface, because it is by definition perpendicular to that surface.
  • The tension in a string does no work on a body moving in a circle, which is why circular motion at constant speed needs no energy input.
  • The magnetic force on a moving charge does no work, being always perpendicular to the velocity — the reason a magnetic field can change a particle's direction but never its speed.

No force. A body coasting in deep space has no work done on it, and its energy is unchanged — the energy statement of Newton's first law.

4. Work Done by a Varying Force: the Area Under the Graph

The formula W=Fscos⁡θW = Fs\cos\theta assumes the force is constant. When it varies — a spring being stretched, a rocket's thrust changing, a bow being drawn — the displacement is split into intervals so short that FF is effectively constant over each, and the contributions are summed:

W=∫F dsW = \int F\,ds

Geometrically, the work done is the area under a force–displacement graph. This is the general definition, of which W=Fscos⁡θW = Fs\cos\theta is the special case where the graph is a horizontal line and the area is a rectangle.

Worked example — stretching a spring. A spring obeying F=kxF = kx is stretched from its natural length to an extension xx. The force–extension graph is a straight line from the origin to (x,kx)(x, kx), so the area is a triangle:

W=12 base×height=12x(kx)=12kx2W = \tfrac{1}{2}\,\text{base}\times\text{height} = \tfrac{1}{2}x(kx) = \tfrac{1}{2}kx^{2}

which is the elastic potential energy stored, met in Block A and now derived. The same integral confirms it:

W=∫0xkx′ dx′=[12kx′2]0x=12kx2W = \int_0^x kx'\,dx' = \left[\tfrac{1}{2}kx'^{2}\right]_0^x = \tfrac{1}{2}kx^{2}

Note the factor of 12\tfrac{1}{2}: using W=Fx=kx2W = Fx = kx^{2} would double-count, because the force is not kxkx throughout the stretch — it grows from zero.

The area method survives even where no formula exists. For a non-Hookean material, or an experimentally measured force curve, the work is found by counting squares under the graph, exactly as in the graph chapter of Block A.

5. Work Done by Gravity and the Path-Independence of Conservative Forces

The work done by gravity on a body of mass mm that descends a vertical height hh is

Wgravity=mghW_{\text{gravity}} = mgh

positive when the body falls and negative when it rises. The remarkable feature is that this depends only on the change in height, not on the path taken. A ball dropped 2 m2~\text{m}, slid down a 2 m2~\text{m}-high frictionless ramp, or carried down a spiral staircase all have exactly mghmgh of work done on them by gravity.

The reason is the second reading of the definition in Section 1: gravity acts vertically, so the only part of the displacement that counts is its vertical component — and the total vertical component of a journey is just the change in height, however winding the route.

A force with this property is called conservative, and the definition can be stated in two equivalent ways:

  • The work done is independent of the path between two points.
  • The work done around any closed loop is zero.

Gravity, the elastic force and the electrostatic force are conservative. Friction and drag are not: dragging a crate around a closed loop back to its starting point does a large amount of negative work, not zero, since friction reverses direction whenever the motion does. This distinction is exactly what makes potential energy definable for gravity and not for friction, as the next chapters develop.

6. A Full Worked Example

Problem. A crate of mass 32 kg32~\text{kg} is dragged 12 m12~\text{m} along a horizontal floor by a rope at 28∘28^\circ above the horizontal with a tension of 145 N145~\text{N}. The coefficient of kinetic friction is 0.240.24. Find the work done by each force and the total work.

Normal force — needed before friction can be computed. Vertically there is no acceleration, and the rope pulls partly upward:

N=mg−Tsin⁡28∘=32(9.81)−145(0.469)=314−68.0=246 NN = mg - T\sin 28^\circ = 32(9.81) - 145(0.469) = 314 - 68.0 = 246~\text{N}

Work done by the tension (θ=28∘\theta = 28^\circ):

WT=Tscos⁡28∘=145(12)(0.883)=1.54×103 JW_T = Ts\cos 28^\circ = 145(12)(0.883) = 1.54\times 10^{3}~\text{J}

Work done by friction (θ=180∘\theta = 180^\circ, so cos⁡θ=−1\cos\theta = -1):

f=μkN=0.24(246)=59.0 NWf=−fs=−59.0(12)=−708 Jf = \mu_k N = 0.24(246) = 59.0~\text{N} \qquad W_f = -fs = -59.0(12) = -708~\text{J}

Work done by the weight and the normal force: both are perpendicular to the horizontal displacement, so

WW=0WN=0W_W = 0 \qquad W_N = 0

Total work:

Wtotal=1540−708+0+0=8.3×102 JW_{\text{total}} = 1540 - 708 + 0 + 0 = 8.3\times 10^{2}~\text{J}

This net positive figure is exactly the kinetic energy the crate gains — the content of the work–energy theorem, which is the subject of the next chapter.

7. Common Exam Pitfalls

  • Omitting the cosine. W=FsW = Fs is valid only when the force is along the displacement.
  • Using the distance travelled instead of the displacement in the force's direction. For a body moving in a circle and returning to its start, the work done by a constant force is zero.
  • Claiming work is done when nothing moves. Holding a weight stationary transfers no energy to it, however great the effort.
  • Forgetting that friction does negative work. Its angle to the motion is 180∘180^\circ, so cos⁡θ=−1\cos\theta = -1.
  • Using W=FxW = Fx for a spring. The force varies, so the work is the triangular area 12kx2\tfrac{1}{2}kx^{2}, not kx2kx^{2}.
  • Assuming the normal force does work. It is perpendicular to the motion along a surface and therefore never does.
  • Treating work as a vector. It is a scalar; its sign indicates the direction of energy flow, not a direction in space.
  • Forgetting to compute NN before friction when the pulling force is angled — the vertical component changes NN and therefore the friction and its work.

Summary

  • Work is the transfer of energy by a force: W=Fscos⁡θ=F⃗⋅s⃗W = Fs\cos\theta = \vec{F}\cdot\vec{s}, measured in joules, where 1 J=1 N⋅m1~\text{J} = 1~\text{N}\cdot\text{m}.
  • The cosine appears because only the component of the force parallel to the displacement transfers energy.
  • Work is a scalar with a sign: positive when the force has a component along the motion (energy in), negative when it opposes the motion (energy out), zero when perpendicular.
  • Zero work occurs when there is no displacement, when the force is perpendicular to the motion (normal force, string tension in circular motion, magnetic force), or when there is no force.
  • For a varying force, W=∫F dsW = \displaystyle\int F\,ds — the area under a force–displacement graph. Stretching a spring gives the triangular area 12kx2\tfrac{1}{2}kx^{2}.
  • Gravity does work mghmgh, depending only on the change in height and not on the path.
  • A conservative force does path-independent work, equivalently zero work around a closed loop. Gravity and elastic forces are conservative; friction and drag are not, which is why potential energy can be defined for the former and not the latter.