Holding a Heavy Bag Still Tires You but Does No Work

Standing still with a heavy school bag for ten minutes is exhausting, yet physics says you did no work on the bag. Work in physics needs a force and a displacement along that force.

This precise idea of work connects directly to kinetic energy, the energy of motion, through one of the most useful results in mechanics.

This part covers work by a constant force, kinetic energy, the work-energy theorem, and forces that do zero work.

**Work done by a constant force is the scalar product of force and displacement, $W=\vec{F}\cdot \vec{d}=Fd\cos \theta$; it is positive for $\theta <90°$, zero at $90°$, and negative for $\theta >90°$.

The unit is the joule**: $1$ J $=1$ N m.

Worked example 1 — at an angle. A $50$ N pull at $60°$ to the ground moves a trolley $10$ m.

$$W=50\times 10\times \cos 60°=250\text{ J}$$

Worked example 2 — vector form. $\vec{F}=4\hat{i}+3\hat{j}$ N and $\vec{d}=2\hat{i}-5\hat{j}$ m.

$$W=(4)(2)+(3)(-5)=-7\text{ J}$$

Worked example 3 — signs. A $2$ kg bag is lifted $3$ m at steady speed, with $g=10$ m/s${}^2$.

- Work by the hand $=+20\times 3=+60$ J (force and displacement both up)
- Work by gravity $=-60$ J (force down, displacement up)

An everyday example. Pulling a trolley bag by its tilted handle, only the component of your pull along the floor does work.

The substance. Negative work removes energy from a body — friction and braking forces do negative work.

**Kinetic energy is the energy a body has because of its motion, $K=\frac{1}{2}m{v}^2$; it is a scalar, never negative, and can also be written $K=\frac{p^2}{2m}$.

Worked example 1 — a car.** A $1000$ kg car at $72$ km/h $=20$ m/s:

$$K=\frac{1}{2}\times 1000\times 20^2=2\times 10^5\text{ J}$$

Worked example 2 — a ball and a bullet.

$$K_{ball}=\frac{1}{2}(0.15)(30{)}^2=67.5\text{ J}{K}_{bullet}=\frac{1}{2}(0.01)(500{)}^2=1250\text{ J}$$

Worked example 3 — speed matters more. Doubling the car's speed to $40$ m/s gives $8\times 10^5$ J, four times as much.

Worked example 4 — same momentum. Bodies of $1$ kg and $5$ kg both have $p=10$ kg m/s:

$$K_1=\frac{100}{2}=50\text{ J}{K}_5=\frac{100}{10}=10\text{ J}$$

For equal momentum, the lighter body has more kinetic energy.

An everyday example. A loaded truck and a scooter at the same speed differ hugely in kinetic energy because of mass, which is why the truck needs far more braking.

The substance. Kinetic energy depends on the frame — a passenger sitting in a moving train has zero kinetic energy relative to the train.

**The work done by the net force on a body equals the change in its kinetic energy, $W_{net}=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$.

Derivation for constant force.** From $v^2-u^2=2as$, multiply by $\frac{m}{2}$:

$$\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=mas=Fs=W_{net}$$

For a varying force, $W=\int Fdx$, the area under the force-position graph; for example, $F=3x^2$ N from $x=0$ to $2$ m gives $W=[x^3{]}_0^2=8$ J.

Worked example 1 — with friction. A $10$ N push moves a $2$ kg block $5$ m from rest against $4$ N of friction.

$$W_{net}=(10-4)\times 5=30\text{ J}=\frac{1}{2}(2)v^2\Rightarrow v=\sqrt{30}\approx 5.48\text{ m/s}$$

Worked example 2 — braking. A $1000$ kg car at $20$ m/s brakes with a $5000$ N force.

$$5000\times d=2\times 10^5\Rightarrow d=40\text{ m}$$

Worked example 3 — a bullet. A $0.01$ kg bullet at $400$ m/s stops inside a plank after $0.05$ m. The average resisting force is

$$F=\frac{\frac{1}{2}(0.01)(400{)}^2}{0.05}=\frac{800}{0.05}=16000\text{ N}$$

An everyday example. A bicycle freewheeling on a flat road slows to a stop as friction and air drag do negative work equal to its kinetic energy.

The substance. Use the net work of all forces, including friction and gravity, not just the applied force.

A force does zero work when it is perpendicular to the displacement, when its point of application does not move, or when there is no displacement at all.

Worked example 1 — circular motion. A stone whirled in a $0.5$ m circle has a string tension of $5$ N pointing to the centre. In each small step the displacement is along the tangent, so

$$W=Fd\cos 90°=0$$

Over a full turn of path length $2\pi (0.5)\approx 3.14$ m, the tension still does zero work, which is why the stone's speed stays constant.

Worked example 2 — no displacement. Pushing a wall with $200$ N, or holding a $20$ kg bag still for ten minutes, does zero mechanical work on the wall or the bag.

An everyday example. A railway porter carrying luggage on the head along a level platform does no work on the luggage against gravity, because the upward support is perpendicular to the horizontal motion.

The substance. Feeling tired is not the same as doing work — your muscles use energy internally without transferring it to the bag.