Work, Density and Center of Mass

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If a force component $F(x)$ acts along a displacement, its work is $W=\int_a^bF(x)dx$ . Force times distance has units of energy. A signed component can produce negative work: friction removes mechanical energy when its direction opposes motion.

A thin rod with linear density $\lambda(x)$ has mass $m=\int_a^b\lambda(x)dx$ . Its moment about the origin is $M=\int_a^bx\lambda(x)dx$ , and its center of mass is $\bar x=M/m$ when $m>0$ . The coordinate is weighted by mass, not merely averaged across the endpoints.

Visual guide

VISUAL GUIDEWork is accumulated force over displacement

Force F = 2x

A spring with k = 2 has force F(x) = 2x. Stretching from 0 to 3 requires the shaded work ∫₀³2x dx = 9. Using only the final force 6 times distance 3 would overestimate the work.

Worked example: stretching a spring

A spring requires $12$ N to stretch $0.03$ m beyond its natural length. Hooke's law $F=kx$ gives $k=400$ N/m. The work to stretch from $0.02$ m to $0.05$ m is

W=\int_{0.02}^{0.05}400x\,dx=200(0.05^2-0.02^2)=0.42\text{ J}.

Using the final force over the entire displacement overestimates the work because the force rises continuously. The model assumes the spring remains within its linear elastic range.

Worked example: a nonuniform rod

For $0\le x\le2$ m and $\lambda(x)=1+x$ kg/m (with $x$ numerically in meters),

m=\int_0^2(1+x)dx=4\text{ kg},

M=\int_0^2x(1+x)dx=\frac{14}{3}\text{ kg m}.

Thus $\bar x=7/6$ m. The center lies to the right of the midpoint because density increases to the right, and it lies inside the rod, as required for nonnegative density.

For pumping problems, a horizontal fluid slice has weight $\rho gA(y)dy$ and lifting distance $h-y$ . Integrate $\rho gA(y)(h-y)dy$ over the initial fluid levels. This assumes constant density, quasistatic lifting and negligible losses; it calculates ideal required work.

Practice

Find work for a constant $5$ N force over $3$ m in its direction.
Find the center of a uniform rod on $[-1,3]$ .
Water fills a tank with constant cross-sectional area $2\text{ m}^2$ from $y=0$ to $1$ m. Set up work to lift it to $y=2$ m.

Show worked solutions

$W=5\cdot3=15$ J.
Constant density cancels: $\bar x=(\int_{-1}^3x dx)/4=1$ .
$W=\int_0^1 1000\cdot9.81\cdot2(2-y)dy=29430$ J under the ideal model.