Work, Energy and Power — Formulas & Examples

Work, energy, and power describe different parts of the same story in physics. Work is energy transferred by a force through a displacement, energy tells you how much can change in a system, and power tells you how fast that transfer happens.

If you only remember one distinction, use this: work is a transfer, energy is an amount, and power is a rate. That single check clears up most beginner mistakes.

Work, energy, and power formulas

For a constant force acting over a displacement,

$$W = \vec{F} \cdot \Delta \vec{r} = Fd\cos\theta$$

Here, $\theta$ is the angle between the force and the displacement. If the force points along the motion, the work is positive. If it points against the motion, the work is negative. If it stays perpendicular to the motion, the work is zero.

In mechanics, two common energy formulas are

$$K = \frac{1}{2}mv^2$$

and, near Earth's surface,

$$\Delta U_g = mg\Delta h$$

The first is kinetic energy. The second is the change in gravitational potential energy. The reference level for potential energy is chosen by you, but the change $\Delta U_g$ is what matters in most problems.

Average power is

$$P_{avg} = \frac{W}{\Delta t}$$

At an instant, the mechanical power delivered by a force is

$$P = \vec{F} \cdot \vec{v}$$

That instantaneous form is useful only when you mean force and velocity at the same moment.

How work, energy, and power fit together

Work is the link between force and energy. If the net work on an object is positive, its kinetic energy increases. If the net work is negative, its kinetic energy decreases.

Power does not tell you how much energy changed by itself. It tells you how quickly the change happened. Two machines can do the same amount of work, but the one that does it in less time has greater power.

In symbols, that kinetic-energy link is the work-energy theorem:

$$W_{net} = \Delta K$$

This is about net work, not just one force acting alone.

Worked example: lifting a backpack

Suppose you lift a $10\,\mathrm{kg}$ backpack straight upward by $2\,\mathrm{m}$ in $4\,\mathrm{s}$ at roughly constant speed.

Because the speed is approximately constant, the upward force you apply is approximately equal to the backpack's weight:

$$F \approx mg = (10)(9.8) = 98\,\mathrm{N}$$

The force and displacement point in the same direction, so $\theta = 0$ and $\cos\theta = 1$. The work you do on the backpack is

$$W_{you} = Fd = (98)(2) = 196\,\mathrm{J}$$

That same lift increases the backpack's gravitational potential energy by

$$\Delta U_g = mg\Delta h = (10)(9.8)(2) = 196\,\mathrm{J}$$

So in this case, the work you do becomes gravitational potential energy.

Now calculate the average power:

$$P_{avg} = \frac{W}{\Delta t} = \frac{196}{4} = 49\,\mathrm{W}$$

The backpack gains $196\,\mathrm{J}$ of gravitational potential energy, and you transfer that energy at an average rate of $49\,\mathrm{J/s}$, which is $49\,\mathrm{W}$.

One subtle point matters here. The net work on the backpack is approximately zero because your positive work is balanced by gravity's negative work. That is consistent with the backpack moving at nearly constant speed, so its kinetic energy does not change much.

Common mistakes in work, energy, and power problems

• Using $W = Fd$ when the force is not along the displacement. If there is an angle, the correct constant-force form is $W = Fd\cos\theta$.
• Confusing energy with power. Energy answers "how much"; power answers "how fast."
• Forgetting the condition behind a formula. For example, $W = Fd\cos\theta$ is the simple form for a constant force, and $\Delta U_g = mg\Delta h$ is the near-Earth approximation.
• Thinking positive work always means speed increases. It is the net work that determines the change in kinetic energy.
• Treating watts and joules as the same unit. A joule is energy; a watt is joule per second.

Where work, energy, and power are used

These ideas show up whenever you care about forces, motion, and energy transfer together. Common cases include lifting objects, braking, motors, climbing stairs, falling bodies, and machine efficiency.

In classroom physics, this topic is also the foundation for the work-energy theorem and for many conservation-of-energy problems. Once you know which quantity a problem is really asking about, the setup usually becomes much shorter.

Quick check: which quantity does the problem want?

Ask these three questions:

1. Is the problem asking how much energy was transferred by a force? Use work.
2. Is it asking about stored or changing motion/position energy? Use an energy equation.
3. Is it asking how quickly the transfer happened? Use power.

That check prevents most mix-ups before the algebra even starts.

Try a similar problem

Try your own version of the backpack example, but lift the same backpack through the same height in $2\,\mathrm{s}$ instead of $4\,\mathrm{s}$. The work and energy change stay the same, but the average power does not. Solve it and compare the result with the original case.