Why a Bucket of Water Swung Overhead Does Not Spill

Stretch a resistance band and the pull grows the farther you stretch — so $W=Fd$ no longer works directly. You need the area under the force-displacement graph.

Some forces, like a spring's pull or gravity, store the work done against them and give it back. That stored energy is potential energy, and it makes energy conservation a powerful shortcut.

This part covers variable forces, conservative forces, spring energy with energy conservation, and motion in a vertical circle.

**Work done by a variable force is the area under its force-displacement graph, which equals the integral $W={\int }_{x_1}^{x_2}F(x)dx$.**

For a constant force, the area is a rectangle, $F\times d$ — the familiar formula is just a special case.

Worked example 1 — from a graph. A force rises steadily from $0$ to $10$ N over the first $2$ m, then stays at $10$ N up to $x=5$ m.

$$W=\underset{\text{triangle}}{\underbrace{\frac{1}{2}(2)(10)}}+\underset{\text{rectangle}}{\underbrace{(3)(10)}}=10+30=40\text{ J}$$

Worked example 2 — by integration. $F=3x^2+2x$ N acts from $x=0$ to $x=2$ m.

$$W={\int }_0^2(3x^2+2x)dx={\left[x^3+x^2\right]}_0^2=8+4=12\text{ J}$$

Area below the displacement axis counts as negative work.

An everyday example. Drawing a bow in archery: the first few centimetres are easy, the last few are hard, and the work done is the area under that rising force curve.

The substance. An F-x graph gives work, while an F-t graph gives impulse — do not mix up the two areas.

**A force is conservative if the work it does depends only on the start and end points, not the path, so the work around any closed path is zero; for such a force, potential energy is defined by $\Delta U=-W_{\text{conservative}}$, and $F=-\frac{dU}{dx}$.

- Conservative: gravity, spring force, electrostatic force
- Non-conservative: friction, air resistance — the work depends on the path and the energy is lost as heat

Worked example 1 — gravity.** Lift a $2$ kg bag $5$ m by stairs or by a long ramp, with $g=10$ m/s${}^2$. Either way, gravity does $-100$ J, so $\Delta U=+100$ J.

Worked example 2 — friction. A box meets $20$ N of friction. Dragging it straight between two points $5$ m apart, friction does $-100$ J; going round by legs of $3$ m and $4$ m, it does $-140$ J. Path matters, so friction has no potential energy.

Worked example 3 — force from energy. If $U=5x^2$ J, then

$$F=-\frac{dU}{dx}=-10x\text{ N}$$

An everyday example. Reaching a hilltop temple by the steps or by the winding road gives you the same gain in gravitational potential energy.

The substance. Only changes in potential energy matter — the zero level can be chosen anywhere.

**Stretching or compressing a spring by $x$ stores $U=\frac{1}{2}k{x}^2$, and when only conservative forces act, kinetic plus potential energy stays constant: $K_1+U_1=K_2+U_2$.

Deriving spring energy.** The spring pulls back with $F=-kx$, so the work done against it is

$$W={\int }_0^{x}kxdx=\frac{1}{2}k{x}^2$$

Worked example 1 — a spring launcher. A spring with $k=200$ N/m is compressed $0.1$ m and launches a $0.05$ kg ball.

$$U=\frac{1}{2}(200)(0.1{)}^2=1\text{ J}=\frac{1}{2}(0.05)v^2\Rightarrow v=\sqrt{40}\approx 6.32\text{ m/s}$$

Worked example 2 — gravity. A ball dropped from $20$ m, with $g=10$ m/s${}^2$:

$$mgh=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{2(10)(20)}=20\text{ m/s at the ground}$$

After falling $5$ m, $v=\sqrt{2(10)(5)}=10$ m/s.

Worked example 3 — maximum compression. A $2$ kg block at $4$ m/s hits a spring with $k=800$ N/m on a smooth floor.

$$\frac{1}{2}(2)(4{)}^2=\frac{1}{2}(800)x^2\Rightarrow x^2=0.04\Rightarrow x=0.2\text{ m}$$

An everyday example. A toy dart gun turns the energy stored in its compressed spring into the dart's kinetic energy.

The substance. If friction acts, mechanical energy falls by exactly the work done against friction.

**For a body on a string of length $r$ to just complete a vertical circle, the tension at the top can drop to zero, so $v_{top}=\sqrt{gr}$, and energy conservation then gives $v_{bottom}=\sqrt{5gr}$.

At the top.** Gravity alone supplies the centripetal force when $T=0$:

$$mg=\frac{m{v}_{top}^2}{r}\Rightarrow v_{top}=\sqrt{gr}$$

From bottom to top the body rises $2r$:

$$\frac{1}{2}m{v}_{bottom}^2=\frac{1}{2}m{v}_{top}^2+mg(2r)\Rightarrow v_{bottom}^2=gr+4gr=5gr$$

Tension at the bottom is then $T=\frac{m{v}^2}{r}+mg=6mg$.

Worked example. A $0.5$ kg stone on a $1$ m string, with $g=10$ m/s${}^2$:

$$v_{top}=\sqrt{10}\approx 3.16\text{ m/s}{v}_{bottom}=\sqrt{50}\approx 7.07\text{ m/s}{T}_{bottom}=6(0.5)(10)=30\text{ N}$$

At the horizontal position, $v^2=5gr-2gr=3gr$, so $v=\sqrt{30}\approx 5.48$ m/s.

An everyday example. A bucket of water whirled overhead does not spill as long as it moves at least $\sqrt{gr}$ at the top.

The substance. **A body on a rigid rod needs only $\sqrt{4gr}$ at the bottom**, because a rod can push as well as pull, so the speed at the top may fall to zero.