Exercises — Work done by variable force — integration

Recognise: $F$ is a single fixed number — it does not depend on $x$. This is the constant-force special case, so no integral is needed (though ${\int }_0^{5}12dx=12\cdot 5$ gives the same thing). $W=Fd=12\times 5=60\text{J}$

• (a) $F$ constant, motion straight → $W=Fd$ legal.
• (b) $F$ depends on $x$ → must integrate.
• (c) $F$ is a constant magnitude, but a bent path means the angle between $\vec{F}$ and motion changes, so you still need $W=\int \vec{F}\cdot d\vec{r}$ — integrate (see Dot product and components of vectors).
• (d) $F$ depends on $x$ → must integrate.

Work = area under the curve = area of the rectangle. $W=10\times 6=60\text{J}$ A rectangle is just the constant-force case drawn as an area — reassuring that both pictures agree.

Why integrate? $F$ grows with $x$, so no single value works. $W={\int }_0^{3}4xdx=4{\left[\frac{x^2}{2}\right]}_0^3=2\left(3^2-0\right)=18\text{J}$ Check by geometry: the graph of $F=4x$ is a straight line from $(0,0)$ to $(3,12)$; a triangle of base $3$, height $12$, area $\frac{1}{2}(3)(12)=18\text{J}$. ✓

$W={\int }_1^2(6x^2-2)dx=[2x^3-2x{]}_1^2$ What we did: the antiderivative of $6x^2$ is $2x^3$ (bump the power up one, divide by the new power), and of $-2$ is $-2x$. $=(2\cdot 8-4)-(2-2)=(16-4)-0=12\text{J}$

$W={\int }_0^4(-3x)dx=-3{\left[\frac{x^2}{2}\right]}_0^4=-3\left(\frac{16}{2}\right)=-24\text{J}$ Sign meaning: every sliver $Fdx$ is negative because the force opposes the displacement — energy is being removed from the body (like a brake). The area lies below the $x$-axis.

Split the area into a triangle and a rectangle.

• Triangle $(0\to 3)$: $\frac{1}{2}(3)(30)=45\text{J}$
• Rectangle $(3\to 7)$: $(7-3)(30)=120\text{J}$ $W=45+120=165\text{J}$

Signed area — the part below the axis subtracts.

• Above axis $(0\to 2)$: $+10\times 2=+20\text{J}$
• Below axis $(2\to 5)$: $-10\times 3=-30\text{J}$ $W_{\text{net}}=20+(-30)=-10\text{J}$ Why it can be negative overall: more energy was taken out (during the opposing stretch) than put in.

$W={\int }_0^4(x-2)dx={\left[\frac{x^2}{2}-2x\right]}_0^4=\left(8-8\right)-0=0\text{J}$ Why zero? The line $F=x-2$ is negative for $x<2$ (force opposes motion) and positive for $x>2$. The triangle below the axis $(0\to 2)$ has area $\frac{1}{2}(2)(2)=2\text{J}$ below, and the triangle above $(2\to 4)$ has area $\frac{1}{2}(2)(2)=2\text{J}$ above. They cancel exactly.

The applied force is $F_{\text{app}}=kx$; the spring force is $F_{\text{spring}}=-kx$ (see Hooke's law and spring potential energy). (a) Work by you: $W_{\text{you}}={\int }_0^{0.1}kxdx=\frac{1}{2}k{x}_0^2=\frac{1}{2}(200)(0.1{)}^2=1\text{J}$ (b) Work by the spring is the negative of that (opposite sign force, same magnitude of stretch): $W_{\text{spring}}=-\frac{1}{2}k{x}_0^2=-1\text{J}$

Step 1 — work: $W={\int }_0^{2}6xdx=3{\left[x^2\right]}_0^2=3(4)=12\text{J}$ Step 2 — work–energy theorem says $W=\Delta KE=\frac{1}{2}m{v}^2-0$: $12=\frac{1}{2}(2)v^2=v^2\Rightarrow v=\sqrt{12}\approx 3.46\text{m/s}$

(b) Reversal point: force is zero when $12-4x=0\Rightarrow x=3\text{m}$. So over $0\to 3$ the force stays positive (only touching zero at the end) — it never actually reverses inside the interval. (a) Net work: $W={\int }_0^3(12-4x)dx={\left[12x-2x^2\right]}_0^3=(36-18)-0=18\text{J}$

(a) $W={\int }_0^2(4-x^2)dx={\left[4x-\frac{x^3}{3}\right]}_0^2=\left(8-\frac{8}{3}\right)-0=\frac{16}{3}\approx 5.33\text{J}$ (b) Work–energy theorem with $v_i=0$: $\frac{16}{3}=\frac{1}{2}(1)v^2\Rightarrow v^2=\frac{32}{3}\Rightarrow v=\sqrt{\frac{32}{3}}\approx 3.27\text{m/s}$ (c) For $x>2$, $x^2>4$, so $F=4-x^2<0$ — the force now points backward, decelerating the particle. At exactly $x=2$ the force is zero (a turning-over point of the push).

$W={\int }_2^{2}5xdx=0\text{J}$ Equal limits ⇒ zero-width interval ⇒ zero area ⇒ zero work. Rule: work needs displacement; with no net movement (in a straight-line, single-valued force), the sum of slivers is empty. (For back-and-forth paths you'd split the integral into legs and add signed pieces — but here the position never changed at all.)

Rewrite $F=6x^{-2}$. Its antiderivative: raise power to $-1$, divide by $-1$: $6\cdot \frac{x^{-1}}{-1}=-\frac{6}{x}$. $W={\int }_1^{3}6x^{-2}dx={\left[-\frac{6}{x}\right]}_1^3=\left(-\frac{6}{3}\right)-\left(-\frac{6}{1}\right)=-2+6=4\text{J}$ Limiting behaviour: if the body went all the way to $x\to \infty$, $W={\left[-\frac{6}{x}\right]}_1^{\infty }=0-(-6)=6\text{J}$ The work stays finite even over an infinite distance, because the force fades as $1/x^2$ — the tail area is small enough to sum to a finite number. This is exactly the kind of reasoning behind escape-energy calculations.