91Ó°ÊÓ

In physics, integrals are used to calculate quantities that accumulate over a range, such as work done, area under curves, or even mass and charge distributions. Specifically, when dealing with the calculation of work done by a force, the integration approach enables us to consider varying force over a range of motion.
To compute work done by a varying force, we integrate the force function concerning the variable over which it acts, typically distance. For example, if a force varies with distance and is given as some function of distance, the work done can be expressed as \[ W = \int F(s) \, ds \]Here, the integration symbol \( \int \) indicates that we sum up the infinitesimal work done, \( F(s) \, ds \), over the entire path.
This approach is beneficial in real-world scenarios where force is not constant, and thus, direct multiplication is insufficient to determine work done.

Inverse proportionality describes a relationship where one quantity increases as the other decreases, such that their product remains constant. Mathematically, if a variable \( y \) is inversely proportional to another variable \( x \), it is expressed as \[ y = \frac{k}{x} \]where \( k \) is a constant.
This relationship is common in physics, such as in gravitational or electrostatic fields, where forces decrease as the distance between interacting bodies increases.

• In the given problem, the force \( F \) varies inversely with distance \( s \), expressed as \( F = \frac{k}{s} \).
• This suggests that as the body moves further away, the force acting on it reduces, holding the product \( F \times s \) constant.

This inverse relationship is crucial in deriving meaningful results when calculating work using integrals.

The concept of how force relates to the distance over which it acts is foundational in understanding motion dynamics. Specifically, this relationship describes how the force applied to an object impacts the movement over distance.
In scenarios where force is constant, the work can be easily explained using the equation \[ W = F \times s \]where \( W \) is the work done, \( F \) is the constant force, and \( s \) is the distance.

• However, when force varies with distance, as explored in the problem, integration is needed due to the changing value of \( F \).
• In the given solution, \( F \) is inversely related to \( s \), a scenario illustrating that the force decreases as distance increases.

In practical terms, this affects how much work is done, necessitating a calculation approach that accommodates such variations, typically involving logarithmic or other complex dependencies not directly aligned with simple power relationships.