Work Done by a Force

Newton’s second law describes how forces change motion locally in time. A second, complementary viewpoint asks how a force acts through a displacement. The scalar quantity that measures this force-displacement transfer is called mechanical work.

For a constant force F acting while a particle undergoes a displacement Δr, the work done by that force is

(1)

Using the geometric definition of the dot product,

(2)

where 𝜃 is the angle between the force and the displacement.

Work is therefore a scalar. Its sign tells whether the force has a component along the displacement or opposite the displacement. A force perpendicular to the displacement does zero work at that instant.

The general definition for a force that may vary in magnitude or direction along a path C is

(3)

This article develops the geometry, sign conventions, units, component form, and basic applications of work. It also distinguishes displacement from path length, which prevents a common mistake: work is not generally force magnitude times distance traveled. The work-energy theorem is derived in M03-02, while variable-force integration is developed in greater detail in M03-03.

1 Force and displacement are both vectors

Suppose a particle moves from position r_i to r_f. Its displacement is

(4)

A force need not point along the displacement. Only the component of the force parallel to the displacement contributes to the work.

Resolve the force into components parallel and perpendicular to the displacement:

(5)

The parallel magnitude is

(6)

Therefore

(7)

The perpendicular component contributes nothing because

(8)

This is the first important geometric lesson: a large force can do little or no work if it is nearly perpendicular to the displacement.

2 Positive, zero, and negative work

The sign of work is set by the angle between the force and displacement:

(9)

Three cases are especially important.

1. If 0 ≤ 𝜃 < 90^∘, then cos 𝜃 > 0 and the force does positive work.
2. If 𝜃 = 90^∘, then cos 𝜃 = 0 and the force does zero work.
3. If 90^∘ < 𝜃 ≤ 180^∘, then cos 𝜃 < 0 and the force does negative work.

A useful interpretation is that positive work corresponds to a force component in the direction of motion, while negative work corresponds to a force component opposite the motion. This interpretation will become precise in M03-02 when work is connected to kinetic energy.

A nonzero force can also do zero work when there is no displacement at all. For example, a person holding a stationary object exerts a force on it, but the mechanical work done on the stationary object is zero because Δr = 0.

3 Component form in Cartesian coordinates

If

(10)

and

(11)

then the dot product gives

(12)

This form is often easier than finding the angle explicitly.

For example, if

(13)

and

(14)

then

(15)

4 Units and dimensions

The SI unit of work is the joule:

(16)

Since

(17)

we have

(18)

Thus the dimensions of work are

(19)

Torque can also be expressed dimensionally as force times distance, but torque and work are different physical quantities. Work is a scalar produced by a dot product, while torque is an axial vector produced by a cross product. For this reason torque is conventionally reported in newton-meters rather than joules.

5 Work by several forces

If several forces act on a particle, each force can do its own work. If the forces are constant over the displacement,

(20)

The total work by all forces is

(21)

Because the dot product is distributive,

(22)

for constant forces acting through the same displacement.

The distinction between work done by one force and net work is essential. Gravity may do positive work while a tension force does negative work, or one force may do zero work while another does nonzero work.

6 Work done by common forces

Weight near Earth’s surface

For a vertical displacement Δy measured positive upward,

(23)

Therefore

(24)

Gravity does positive work when the particle moves downward and negative work when it moves upward. The potential energy interpretation is developed later in M03-05 and M03-06.

Normal force

For motion along a fixed smooth surface, the Normal force is perpendicular to the instantaneous displacement tangent to the surface. Thus

(25)

and the normal force does zero work.

This statement has conditions. A normal force can do work when the constraining surface itself moves. The correct test is always the dot product between the force and the displacement of the point on which that force acts.

Kinetic friction

If an object slides across a stationary surface and kinetic friction points opposite the displacement, then

(26)

For a simple horizontal slide with constant f_k = μ_kN,

(27)

The minus sign is a consequence of the force-displacement angle, not a special rule attached to friction. In more complicated systems, including moving surfaces, the work of friction must be evaluated from the actual force and displacement.

Tension

Tension can do positive, negative, or zero work depending on the geometry and which body is being analyzed. For example, tension on a mass being lifted upward can do positive work. The same tension on another connected body moving downward can do negative work.

7 Worked example 1: constant force at an angle

A worker pulls a crate through a horizontal displacement of 12.0 m using a constant force of magnitude 85.0 N directed 30.0^∘ above the horizontal. Find the work done by the applied force.

The displacement is horizontal, so the angle between force and displacement is 30.0^∘:

(28)

Thus

(29)

Therefore

(30)

The vertical component of the applied force does no work because the crate has no vertical displacement.

8 Worked example 2: work by individual forces on a sliding block

A 6.00 kg block slides 5.00 m to the right across a horizontal floor. A horizontal applied force of 32.0 N acts to the right. The kinetic friction coefficient is μ_k = 0.250. Find the work done by the applied force, friction, gravity, and the normal force.

The normal force is

(31)

The kinetic friction magnitude is

(32)

Applied-force work:

(33)

Friction work:

(34)

Gravity and the normal force are perpendicular to the horizontal displacement, so

(35)

Hence

(36)

The net work is therefore 86.4 J. M03-02 will show what this net work implies about the change in speed.

9 Worked example 3: a force perpendicular to circular motion

A particle moves at constant radius R around a circle while a purely radial force always points toward the center. Find the work done by the radial force over any finite arc of the circular path.

For an infinitesimal displacement along the circle, dr is tangent to the path, while the radial force is perpendicular to the tangent. Therefore

(37)

Integrating around any arc,

(38)

A force can continuously change the direction of velocity while doing zero work. This is the force counterpart of uniform circular motion developed in M01-08 and M02-11.

10 Displacement, distance, and path

For a constant force, the work can be written as F ⋅ Δr, so the endpoint displacement is the relevant vector. It is generally incorrect to replace Δr by the total distance traveled unless the force remains tangent to the motion with the appropriate sign.

For example, if a particle moves out and then returns to its starting point while a single constant force acts throughout, the total displacement is zero. The total work by that constant force over the complete round trip is therefore zero, even though the particle traveled a nonzero distance.

For a force that changes with position or direction, the actual path can matter. This is why the general definition uses a line integral along the path C.

11 The force versus position graph

For one-dimensional motion along x, the differential work is

(39)

For a constant force from x₁ to x₂,

(40)

Geometrically, this is the signed area under the F_x versus x graph.

If F_x varies with position, the rectangle is replaced by the limiting sum of many narrow strips. This leads to

(41)

which is developed systematically in M03-03.

12 Worked example 4: reading work from a force versus position graph

A force along the x direction has constant value F_x = 18.0 N from x = 2.00 m to x = 7.50 m. Find the work done by the force.

The displacement is

(42)

The work is the rectangular area under the graph:

(43)

Thus

(44)

If the same force had been −18.0 N over the same positive displacement, the work would have been −99.0 J.

13 Work along a curved path

For a general path, the displacement changes direction continuously. Divide the path into many small displacement vectors Δr_i. Over a sufficiently small segment, the force can be treated as approximately constant, so

(45)

Summing and taking the limit gives the line integral

(46)

If e_t is the unit tangent and ds is an infinitesimal path length, then

(47)

Therefore

(48)

where F_t is the tangential component of the force. Thus even on a curved path, only the component of force tangent to the path contributes to work.

14 Work depends on the reference frame

Displacement and velocity depend on the reference frame, so mechanical work generally does also. A force can do different amounts of work when the same physical process is described from different inertial frames.

This does not create a contradiction. Kinetic energy is also frame dependent, and the work-energy theorem remains consistent within each frame. The force, displacement, and energy quantities must all be evaluated in the same frame.

15 Common mistakes

1. Using W = Fd without checking the angle between force and displacement.
2. Treating work as a vector because force and displacement are vectors.
3. Assuming every force acting on a moving object does nonzero work.
4. Calling FΔr the work when the force is perpendicular to the displacement.
5. Forgetting that negative work is physically meaningful.
6. Adding a separate “work force” or “centripetal work” to the force diagram.
7. Confusing work done by one force with net work by all forces.
8. Assuming the normal force always does zero work without checking whether the constraint is moving.
9. Assuming friction always does negative work without examining the actual force and displacement.
10. Confusing the joule with the newton. Work has units of force times distance.

16 Practice exercises

1. A 25 N horizontal force moves a box 6.0 m horizontally in the same direction. Find the work done by the force.
2. A 40 N force acts through a displacement of 3.0 m at an angle of 60^∘ to the displacement. Find the work.
3. A force of magnitude 50 N acts perpendicular to a displacement of 8.0 m. Find the work done by that force.
4. A 12 N force acts opposite a 5.0 m displacement. Find the work and state its sign.
5. Evaluate the work for
   
6. A 10 kg crate is raised vertically by 2.5 m. Find the work done by gravity using g = 9.81 m∕s².
7. A block slides 4.0 m across a level floor with kinetic friction magnitude 18 N. Find the work done by friction.
8. A particle moves through a quarter circle while a force always points radially toward the circle center. What work does that radial force do? Explain geometrically.
9. A constant force F_x = −7.0 N acts while a particle moves from x = −2.0 m to x = 5.0 m. Find the work.
10. A force versus position graph is a rectangle of height 14 N from x = 1 m to x = 9 m. Find the work from the signed area. Then state what changes if the rectangle lies below the x axis instead.

17 Answers to practice exercises

1. 150 J.
2. 60 J.
3. 0 J.
4. −60 J.
5. 2 J.
6. −245 J to three significant figures.
7. −72 J.
8. Zero; the radial force is perpendicular to every infinitesimal tangential displacement.
9. −49 J.
10. 112 J; below the axis the signed area and work are −112 J.

18 What comes next

This article defines work geometrically and operationally. M03-02 uses Newton’s second law to derive the work-energy theorem,

(49)

which explains how net work changes a particle’s speed. M03-03 then develops variable-force work and force-position integration in greater detail.

References