Work of a Variable Force

U_{1 \to 2} = \int_{s_{1}}^{s_{2}} F_{t} d s = \int F \cdot d r

Only the tangential component does work. A plot of F vs s: work is the area. Springs (½k(x²) difference) and gravity (mgΔh) are the usual closed forms.

$F_{t}$: Force along the path
$s$: Path coordinate

Step-by-step solved example

F=8−2s (N, m) from s=0 to s=3 m. Work?

Work is the area under F–s (variable force).

1. Integrate
U=∫(8−2s)ds = 8s−s² from 0 to 3=24−9=15 J.
$U = 15 J$
2. Area check
Trapezoid (8+2)/2 × 3=15 J.

Answer: 15 J

10 practice questions

0/10 correct

1.A force perpendicular to the path does

2.Work of a spring between x1 and x2 is

3.Constant F=12 N over 4 m along F: U=

4.Work–energy: U_all =

5.Power P=

6.Negative work means

7.F=2s from 1 to 3 m: U=

8.Gravity work depends only on

9.Kinetic friction work is

10.If F vs s is a triangle 0–10 N–0 over 4 m, U=