5. Work Vs Force Modeller — Theory
Work is energy transferred when a force acts through a displacement. This page explains the integral setups used by the calculator.
1) Work from a variable force
\[
W = \int_a^b F(x)\,dx
\]
If the force depends on position, you can’t use \(W=F\Delta x\) with a single constant \(F\).
Instead, you slice the motion into tiny pieces of length \(dx\). Over each tiny piece the force is approximately constant:
\(dW \approx F(x)\,dx\). Summing and taking the limit gives the integral above.
Sign convention: If \(F(x)\) is negative on part of the interval, the integral can be negative.
That corresponds to the force removing energy (doing negative work) over that region.
Units: \(F\) in newtons and \(x\) in meters gives
\(\mathrm{N}\cdot\mathrm{m}=\mathrm{J}\) (joules).
2) Hooke’s law (spring work)
\[
F(x)=kx
\qquad\Longrightarrow\qquad
W=\int_{x=a}^{x=b} kx\,dx
\]
For an ideal spring (measured from its relaxed length), the restoring force magnitude is \(kx\).
The work required to stretch or compress from \(a\) to \(b\) is:
\[
W = k\int_a^b x\,dx
= k\left[\frac{x^2}{2}\right]_a^b
= \frac{k}{2}\left(b^2-a^2\right)
\]
This equals the change in spring potential energy:
\(\Delta U = \frac{1}{2}k(b^2-a^2)\).
3) Pumping water from a tank (slices method)
Suppose the tank is filled up to height \(h\). A horizontal slice at height \(y\) has thickness \(dy\)
and cross-section area \(A(y)\). Its volume is \(A(y)\,dy\).
\[
\text{mass of slice} = \rho\,A(y)\,dy
\qquad\Rightarrow\qquad
\text{weight} = \rho g\,A(y)\,dy
\]
That slice must be lifted distance \(h-y\) to reach the top, so:
\[
dW = (\rho g\,A(y)\,dy)\,(h-y)
= \rho g\,A(y)(h-y)\,dy
\]
\[
\boxed{
W = \rho g \int_0^h A(y)(h-y)\,dy
}
\]
If \(A(y)\) is constant (a cylinder), the integral reduces to a simple polynomial integral.
4) Graph interpretation
In force mode, the graph plots \(F(x)\) versus \(x\) and shades the region under the curve on \([a,b]\).
The signed area equals the work.
In pumping mode, the graph plots the integrand \(\rho g A(y)(h-y)\) versus \(y\) and shades \([0,h]\).
The shaded area equals the pumping work.
5) Tips and common pitfalls
- Keep units consistent. If you input distances in cm or ft, the calculator converts to meters internally.
- Domain issues (like \(\sqrt{x}\) with negative \(x\)) can make the integral fail numerically.
- Negative areas are physically meaningful: they indicate negative work (force opposite displacement).
- Pumping setups depend on where \(y=0\) is measured (bottom vs top). This calculator uses \(y=0\) at the bottom and pumps to height \(h\).
