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Work done by constant and variable forces

Understanding Work: The Basic Concept

In physics, "work" is a very specific term. It's not just about effort or exertion; it's about a force causing an object to move over a distance. For work to be done, two conditions must be met:

A force must act on an object.
The object must move (displace) in the direction of, or at least partially in the direction of, the applied force.

Work is a scalar quantity, meaning it has magnitude but no direction. Its standard unit in the International System of Units (SI) is the joule (J). One joule of work is done when a force of one newton (N) moves an object over a distance of one meter (m) in the direction of the force.

Work Done by a Constant Force

When a constant force acts on an object and causes a displacement, the work done is simply the product of the magnitude of the force and the magnitude of the displacement in the direction of the force.

Consider a constant force F acting on an object that undergoes a displacement d. The angle between the force vector and the displacement vector is denoted by θ.

The work done (W) by this constant force is given by the dot product of the force and displacement vectors:

W = F ⋅ d

This can also be expressed in terms of the magnitudes of the force and displacement, and the angle between them:

W = |F| |d| cos(θ)

Or more simply:

W = Fd cos(θ)

Cases for Constant Force:

When the force is in the same direction as the displacement (θ = 0°): In this case, cos(0°) = 1. So, W = Fd. The work done is positive.
Example: Pushing a box across a floor in the direction you are pushing it.
When the force is in the opposite direction to the displacement (θ = 180°): In this case, cos(180°) = -1. So, W = -Fd. The work done is negative. This means the force opposes the motion.
Example: The force of friction acting on a sliding object.
When the force is perpendicular to the displacement (θ = 90°): In this case, cos(90°) = 0. So, W = 0. The work done is zero. The force does not contribute to the motion.
Example: The centripetal force acting on an object moving in a uniform circular path. The force is always directed towards the center, while the displacement is tangential.
When the force has a component in the direction of displacement: If the force is applied at an angle θ to the displacement, only the component of the force parallel to the displacement (F cos(θ)) does work. The component perpendicular to the displacement (F sin(θ)) does no work.

Work Done by Multiple Constant Forces:

If several constant forces act on an object, the total work done is the algebraic sum of the work done by each individual force.

W_total = W₁ + W₂ + W₃ + ...

Alternatively, the total work done is equal to the work done by the net force (resultant force).

W_total = F_net ⋅ d

Work Done by a Variable Force

In many real-world scenarios, the force acting on an object is not constant. It might change in magnitude, direction, or both as the object moves. When dealing with a variable force, we cannot simply use the W = Fd cos(θ) formula directly. Instead, we must use calculus.

Imagine a force F(x) that varies with position x. If the object moves from position x₁ to x₂ along a certain path, the work done by this variable force is found by integrating the force over the displacement.

Consider the object moving along the x-axis. The force is F(x) = F_x(x) i + F_y(x) j + F_z(x) k. The displacement is dr = dx i + dy j + dz k.

The infinitesimal work done (dW) by the force F over an infinitesimal displacement dr is given by:

dW = F ⋅ dr

To find the total work done (W) as the object moves from an initial position r₁ to a final position r₂, we integrate this expression:

W = ∫_r₁^r₂ F ⋅ dr

Work Done by a Variable Force Along a Straight Line (1D):

If the motion is restricted to a straight line, say the x-axis, and the force F_x(x) depends only on x, then F = F_x(x) i and dr = dx i. The integral simplifies to:

W = ∫_x₁^x₂ F_x(x) dx

Geometrically, this integral represents the area under the curve of the force (F_x) versus displacement (x) graph from x₁ to x₂.

If the F_x vs x graph is given, the work done is the area between the curve and the x-axis. Areas above the x-axis contribute positive work, and areas below the x-axis contribute negative work.

Example: Work done by a spring

A classic example of variable force is the force exerted by a spring, described by Hooke's Law. The force exerted by a spring is F_s = -kx, where k is the spring constant and x is the displacement from the equilibrium position.

To stretch or compress a spring from an initial position x₁ to a final position x₂, the work done by the spring force is:

W_spring = ∫_x₁^x₂ (-kx) dx

W_spring = -k ∫_x₁^x₂ x dx

W_spring = -k [x²/2]_x₁^x₂

W_spring = -k (x₂²/2 - x₁²/2)

W_spring = ½ k x₁² - ½ k x₂²

Note that this is the work done *by the spring*. The work done *by an external agent* to stretch or compress the spring is the negative of this:

W_external = -W_spring = ½ k x₂² - ½ k x₁²

If the spring is stretched from its equilibrium position (x₁ = 0) to a displacement x₂ = x, then W_external = ½ kx².

Key Takeaway: Work is force applied over a distance. For constant forces, W = Fd cos(θ). For variable forces, work is the integral of force over displacement, W = ∫ F ⋅ dr. The graphical interpretation of work done by a 1D variable force is the area under the F-x curve.

Graphical Method for Variable Force (1D)

When a force varies with position along a single axis (e.g., the x-axis), we can visualize the work done by plotting the force F_x as a function of the position x. The work done by the force as the object moves from x₁ to x₂ is precisely the area enclosed between the force-displacement curve and the x-axis, from x₁ to x₂.

Steps to calculate work from a graph:

Identify the initial position (x₁) and the final position (x₂) on the x-axis.
Determine the shape(s) formed between the force-displacement curve and the x-axis within the interval [x₁, x₂]. These shapes are typically rectangles, triangles, or trapezoids.
Calculate the area of each shape. Areas above the x-axis are considered positive contributions to work, while areas below the x-axis are negative contributions.
Sum up the areas of all shapes to find the total work done.

Example: Suppose a force F_x varies linearly from 10 N at x = 0 m to 20 N at x = 2 m. We want to find the work done from x = 0 m to x = 2 m.

The graph of F_x vs x is a straight line. The area under this curve is a trapezoid.

The parallel sides of the trapezoid are the forces at x=0 (F₁ = 10 N) and x=2 (F₂ = 20 N). The height of the trapezoid is the displacement along the x-axis (Δx = 2 m - 0 m = 2 m).

Area = ½ × (sum of parallel sides) × height

Work (W) = ½ × (F₁ + F₂) × Δx

W = ½ × (10 N + 20 N) × 2 m

W = ½ × (30 N) × 2 m

W = 30 Joules

If the force were to become negative in some interval, the area below the x-axis would be subtracted.

Work-Energy Theorem

The Work-Energy Theorem provides a fundamental link between the work done on an object and its kinetic energy. It states that the net work done on an object is equal to the change in its kinetic energy.

Kinetic energy (KE) is the energy an object possesses due to its motion. It is given by the formula:

KE = ½ mv²

where m is the mass of the object and v is its speed.

The Work-Energy Theorem is expressed as:

W_net = ΔKE

W_net = KE_f - KE_i

W_net = ½ mv_f² - ½ mv_i²

Here, W_net is the net work done by all forces acting on the object, v_i is the initial speed, and v_f is the final speed.

This theorem is very powerful because it allows us to find the change in speed of an object if we know the net work done on it, or vice versa, without needing to know the details of the forces or the time taken.

Example: A 2 kg block is initially moving at 5 m/s. A net force does 50 J of work on the block. What is its final speed?

Initial KE = ½ × 2 kg × (5 m/s)² = 25 J

W_net = ΔKE

50 J = KE_f - 25 J

KE_f = 50 J + 25 J = 75 J

½ mv_f² = 75 J

½ × 2 kg × v_f² = 75 J

v_f² = 75 J / 1 kg = 75 m²/s²

v_f = √75 m/s = 5√3 m/s ≈ 8.66 m/s

Conservative and Non-Conservative Forces

Forces can be broadly classified into two types based on the work they do: conservative and non-conservative. This classification is crucial for understanding potential energy.

Conservative Forces:

A force is conservative if the work done by it in moving an object between two points is independent of the path taken between those points. Equivalently, the work done by a conservative force around any closed path is zero.

Key characteristics:

Work done depends only on the initial and final positions.
Work done is zero over a closed path.
Associated with potential energy.

Examples:

Gravitational force
Elastic spring force
Electrostatic force

For a conservative force F, we can define a potential energy function U such that the work done by the force is equal to the negative change in potential energy:

W_conservative = -ΔU = U_initial - U_final

Non-Conservative Forces:

A force is non-conservative if the work done by it depends on the path taken between two points. The work done by a non-conservative force around a closed path is generally not zero.

Key characteristics:

Work done depends on the path taken.
Work done is generally not zero over a closed path.
These forces often dissipate energy, usually as heat or sound.

Examples:

Frictional force
Air resistance
Tension in a string (in some cases)
Applied forces that are not part of a potential energy system

For systems involving both conservative and non-conservative forces, the Work-Energy Theorem can be extended:

W_net = W_conservative + W_{non-conservative} = ΔKE

Substituting W_conservative = -ΔU:

-ΔU + W_{non-conservative} = ΔKE

Rearranging, we get:

W_{non-conservative} = ΔKE + ΔU = Δ(KE + U)

Since E = KE + U is the total mechanical energy, this equation shows that non-conservative forces change the total mechanical energy of a system. If W_{non-conservative} is positive, the total energy increases; if it's negative, the total energy decreases.

Conservation of Mechanical Energy: If only conservative forces do work on a system, the total mechanical energy (KE + U) remains constant. W_{non-conservative} = 0, so Δ(KE + U) = 0, which means KE_initial + U_initial = KE_final + U_final.

Summary of Work Concepts

Work is a fundamental concept linking force, displacement, and energy transfer.

Concept	Formula/Definition	Notes
Work by Constant Force	W = Fd cos(θ)	θ is the angle between F and d. Unit: Joule (J).
Work by Variable Force (1D)	W = ∫_x₁^x₂ F_x(x) dx	Integral of force over displacement. Geometrically, area under F-x curve.
Work by Variable Force (3D)	W = ∫_r₁^r₂ F ⋅ dr	Dot product integration.
Work-Energy Theorem	W_net = ΔKE = ½ mv_f² - ½ mv_i²	Net work done equals change in kinetic energy.
Work by Conservative Force	W_cons = -ΔU	Independent of path, zero over closed loop.
Work by Non-Conservative Force	W_non-cons = ΔKE + ΔU = ΔE_mech	Depends on path, changes total mechanical energy.

Exam Tip: Always identify the forces acting on the object. Determine if they are constant or variable, and conservative or non-conservative. This will guide you to the correct approach (direct calculation, integration, or work-energy theorem). For variable forces, remember the graphical method (area under the curve) if the force is along a single axis.