38. Work and `Definite integral`¹

Dated: 30-10-2024

Work Done by a Constant Force

If a body is moved unit distance \(d\) by a force \(\vec{F}\) in \(x^+\) direction then work done on the body can be found out by

\[W = \vec{F} \cdot d\]

Units

Distance

meter \(m\)

Force

newton \(N\)
pounds \(lb\)
dynes (the force required to accelerate a 1 gram body by \(\frac{1 cm}{s^2}\))

Work

newton meter (also called joule \(j\)) \(N m\)
foot pounds \(ft \, lb\)
dyne centi meters \(dyne \, cm\)

Work Done by a Variable Force

Imagine a body moving in \(x^+\) direction by a force which is dependent on the distance \(x\) in interval² \([a, b]\) then the find the work done on the body.
The work done at any instance would be

\[W_i = F(x_i^*)\Delta x_i\]

The work done across the whole interval² would be

\[\sum_{i=1}^{n}W_{i}=\sum_{i=1}^{n}F(x_{i}^{*})\Delta x_{i}\]

And for more accuracy

\[W=\lim_{\max(\Delta x_{i}\to 0)}\sum_{i=1}^{n}F(x_{i}^{*})\Delta x_{i}=\int_{a}^{b}F(x)dx\]

Pressure

It is defined as force per unit area.

\[P = \frac{force}{area} = \rho \, h\]

Where \(\rho\) is the density and \(h\) is the depth.

Fluid Pressure

If a flat surface is submerged in a fluid at certain depth \(h\) then the pressure on that surface would be

\[F = \rho \cdot h \cdot A\]

The shape of the container is irrelevant.

Pascal's Principle

The fluid pressure at a given height is same in all the directions.

The formula for fluid pressure on a surface would be

\[F = \int_{a}^{b} \rho h(x) w(x) dx\]

38. Work and Definite integral1