Work, Conservative Forces, and Universal Gravitation

A horizontal spring resting on a frictionless table provides an ideal physical model to analyze potential energy without the confounding influence of gravitational potential.
Because the system remains at a uniform horizontal height, gravity acts strictly perpendicular to the direction of motion and does zero work on the mass ( $W_{\text{gravity}} = 0$ ).
The initial mechanical state consists of a box attached to a spring at rest, with the spring resting at its equilibrium length.
An external force (applied by a hand) pushes the box to compress (squish) the spring.
During compression, the spring exerts a restoring force directed opposite to the displacement of the mass, meaning the spring performs negative work ( $W_{\text{spring}} < 0$ ).
Simultaneously, the hand applies a force in the direction of displacement, meaning the hand performs positive work ( $W_{\text{hand}} > 0$ ).

Compressing a spring forces it into a state against its natural equilibrium, causing the potential energy of the spring to increase as it is progressively compressed.
Conservative forces possess an intrinsic "memory" of spatial configuration:
- The gravitational force depends directly on an object's vertical elevation.
- The spring force depends directly on the precise magnitude of elongation or compression relative to equilibrium.
A conservative force or conservative system is defined by its state-dependent nature: the energy of the system depends exclusively on position/spatial coordinates rather than the path taken.

When an object begins at rest and terminates at rest, its initial kinetic energy ( $K_i$ ) and final kinetic energy ( $K_f$ ) are both zero:
- $K_i = 0$
- $K_f = 0$
- $\Delta K = K_f - K_i = 0$
According to the work-energy theorem, the net (total) work performed on the mass must equal the change in kinetic energy:
- $W_{\text{net}} = \Delta K = 0$
The net work of zero arises from the algebraic sum of the positive work done by the external agent and the negative work done by the conservative spring force:
- $W_{\text{net}} = W_{\text{hand}} + W_{\text{spring}} = 0$
A net work value of zero does not signify that no physical changes occurred within the system:
- An external agent expends chemical/metabolic energy (e.g., burning $0.5\text{ calories}$ , with $0.005\text{ calories}$ transferred directly into the compression).
- The spring simultaneously stores that exact quantity ( $0.005\text{ calories}$ ) as elastic potential energy.
Upon releasing the external restraint, the stored potential energy immediately converts into kinetic energy, propelling the mass back toward equilibrium.
A pervasive conceptual misconception regarding work is failing to account for all acting forces:
- Total work must incorporate both conservative and non-conservative forces, external agent inputs, and passive force responses (such as gravity or spring resistance).
- Human intuition tends to register personal physical exertion while overlooking the work executed by passive natural forces.

The elementary expression for gravitational force near the surface of the Earth is:
- $F_g = m \times g$
At planetary, interplanetary, and large orbital scales (such as satellites, the Moon, or the Earth-Sun system), gravity must be described using Newton's Law of Universal Gravitation:
- $F_g = -\frac{G \times M \times m}{r^2}$
- $G$ represents the universal gravitational constant.
- $M$ represents the mass of the primary body (e.g., Earth).
- $m$ represents the mass of the secondary body.
- $r$ represents the radial separation distance measured from the center of the mass.
Evaluating the planetary surface parameter using Earth's mass and radius yields the local gravitational acceleration:
- $\frac{G \times M}{r^2} = 9.8\,m/s^2$
The familiar parameter $g$ in $F_g = m \times g$ is an approximation valid only when vertical displacement variations are negligible compared to the total distance from the center of the Earth.

Work performed by a radially directed variable force over a distance is given by the definite spatial integral:
- $W = \int F(r)\,dr$
Factoring out constant quantities ( $-G \times M \times m$ ):
- $W = -G \times M \times m \int \frac{1}{r^2}\,dr = -G \times M \times m \int r^{-2}\,dr$
Evaluating the power-rule integral:
- $\int r^{-2}\,dr = -r^{-1} = -\frac{1}{r}$
Multiplying by the leading constants cancels the negative signs:
- $W = -G \times M \times m \times \left(-\frac{1}{r}\right) = \frac{G \times M \times m}{r}$

Historical examination assessments generally evaluate universal gravitation through conceptual multiple-choice questions rather than extensive multi-step computational problems.
The derivation demonstrates that gravitational physics extends beyond the localized constant-field equation ( $F_g = m \times g$ ) to accommodate arbitrary spatial scales.

Prompt: As the spring is squished, does the spring do positive work or negative work?
- Response: Negative work.
Prompt: Is the potential energy of the spring increasing or decreasing as it is pushed?
- Response: Increasing (the spring gets progressively "angrier").
Prompt: In the horizontal spring compression scenario, which entity performs positive work?
- Response: The hand.
Prompt: If a mass begins at rest and ends at rest after being pushed against a spring, what is the total (net) amount of work done on the mass?
- Response: Zero.
Prompt: Why is the total work done equal to zero?
- Response: Because the kinetic energy begins at zero and ends at zero ( $\Delta K = 0$ ), meaning positive work performed by the hand cancels out the negative work done by the spring.
Prompt: What is the evaluation of the integral $\int \frac{1}{r^2}\,dr$ (or $\int r^{-2}\,dr$ )?
- Response: $-r^{-1}$ (or $-\frac{1}{r}$ ).