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Work
Results from the application of force over distance. When work is
done, energy is transformed from one form to another.
$$W = Fd$$
where:
Work is measured in joules (J).
Lifting a bag onto a shelf transfers energy from your muscles to the bag.
Pushing sideways on a wall does not transfer energy because the wall does not move.
Thinking that holding a heavy object still means work is being done in physics.
Using a ramp makes lifting a heavy object easier, but the distance moved increases.
Power
Power is the rate at which energy is transferred or work is done.
The defining equation is
$$P = \frac{E}{t}$$
where:
Since $1\ \text{W} = 1\ \text{J s}^{-1}$, a device rated at 1000 W transfers 1000 J of energy every second.
If the energy transfer is mechanical work, then
$$P = \frac{W}{t}$$
A useful rearrangement for constant speed motion is:
$$P = Fv$$
because if $W = Fd$ and $d = vt$, then $P = \frac{Fd}{t} = Fv$.
When you are given a "doing something faster" situation (running upstairs quickly, lifting rapidly, accelerating hard), power is often the target quantity.
An 80 kg person runs up a flight of stairs with a vertical height of 12 m in 5 s.
a) Calculate the energy transferred by the person.
b) Calculate the person’s average power output.
Solution
Working (guided):
Energy transferred:
$$ E = mgh = 80 \times 9.8 \times 12 \approx 9.4 \times 10^3 \text{ J}$$
Average power:
$$P = \frac{E}{t} = \frac{9400}{5} \approx 1900 \text{ W}$$
So the person’s average power output is about 1900 W.
Efficiency
Using scarce resources in the best possible way to avoid welfare loss.
Using energy:
$$\text{efficiency} = \frac{\text{useful output energy}}{\text{total input energy}}\times 100\%$$
Using power:
$$\text{efficiency} = \frac{\text{useful output power}}{\text{total input power}}\times 100\%$$
These are equivalent when the input and output are measured over the same time interval.
A lift is used to raise a mass of 800 kg through a vertical height of 10 m.
The energy supplied to the lift is 47 kJ.
a) Calculate the useful output energy.
b) Calculate the efficiency of the lift.
Solution
Useful output energy:
$$E_\text{useful} = mgh = 800 \times 9.8 \times 10 = 78\,400 \text{ J}$$
Efficiency:
$$\eta = \frac{78\,400}{47\,000} \times 100\% \approx 167\%$$
A filament bulb might be about 5% efficient for visible light output.
If it takes 60 W electrical input, then useful light power is approximately:
$$P_\text{useful} = 0.05\times 60 = 3\ \text{W}$$
Wasted thermal power is:
$$P_\text{wasted} = 60 - 3 = 57\ \text{W}$$
1. Identify whether the question is about energy/work, power, or efficiency (sometimes all three).
2. Write the relevant relationship first: $W=Fd$, $P=E/t$, $P=VI$, or $\eta=\text{useful}/\text{total}$.
3. Track units carefully: J, W, s, N, m, V, A. Convert kJ to J when needed.
4. For efficiency, state clearly what counts as "useful output" in the context.
5. Do a reasonableness check: efficiencies must be between 0% and 100%, and higher efficiency should mean less wasted energy.