Energy, power and efficiency

1. 1If a load is lifted, find its weight first: $F = mg$. This is not on the formula sheet.
2. 2Find the work done with $W = Fs$, using the distance moved in the direction of the force.
3. 3Find the power with $P = W \div t$, with $t$ in seconds. If a steady speed is given instead of a time, use $P = Fv$, which is also not on the sheet.
4. 4Give the answer in joules or watts, to the nearest whole unit unless the question asks for something else.

1. 1Find the work first: $W = P \times t$.
2. 2If the load rises at constant velocity, the work done equals the potential energy gained: $W = mgh$.
3. 3Rearrange for the unknown: $m = W \div (gh)$ or $h = W \div (mg)$.

2020 Question 19a. $W = 4355 \times 90 = 391\,950$ J. The bin rises at constant velocity, so $391\,950 = m \times 9.8 \times 20$, which gives $m = 1999.74 \approx 2000$ kg (the mark guide prints 1999.75).

1. 1Use the energy you were given to recover the missing quantity. From a kinetic energy, $m = 2\,\text{KE} \div v^2$.
2. 2Find the other energy with $\text{PE} = mgh$ or $\text{KE} = \tfrac{1}{2}mv^2$.
3. 3Add them: $\text{ME} = \text{KE} + \text{PE}$. This is not printed anywhere in the formula book.
4. 4For a fall or a launch with no friction, the kinetic energy gained equals the potential energy lost: $mgh = \tfrac{1}{2}mv^2$, so $v = \sqrt{2gh}$ and the mass cancels.

2023 Question 16. $m = 2 \times 850 \div 1.5^2 = 755.56$ kg, then $\text{PE} = 755.56 \times 9.8 \times 2 = 14\,808.98$ J, then $\text{ME} = 850 + 14\,808.98 = 15\,659$ J.

The syllabus names ME = KE + PE in words, and that is all the notice you get. Two past questions turn on it, 2023 Q16 and 2024 Q21, and in 2023 Q16 it carried the final mark.

The motion equations also give the speed of a falling object, and the mark guide accepts both routes. The energy route is shorter and needs no acceleration.

1. 1Decide the direction. The output is always smaller than the input.
2. 2Forwards, with the input known and the output wanted, multiply by the efficiency.
3. 3Backwards, with the output known and the input wanted, divide by the efficiency. For an 80% efficient machine, divide by 0.8. Never multiply by 1.2.
4. 4Check that the input is bigger than the output. If your input comes out smaller than your output, you multiplied when you should have divided. Backwards is always input = output ÷ η.

Two machines, two efficiencies

When one machine drives another, the output of the first is the input of the second. Multiply the efficiencies, for example $0.9 \times 0.8 = 0.72$. The overall efficiency is lower than either one, and this rule is not on the formula sheet.

When each stage takes its own power, as the two pumps do in 2023 Q21, do not multiply. Work out each stage separately, divide each output by its own efficiency, and add the inputs.

1. 1Decide series or parallel by asking what is shared. A series circuit has one path, so the current is the same everywhere. A parallel circuit has branches, so the voltage is the same across every branch.
2. 2Find the total resistance. In series, $R_T = R_1 + R_2 + \dots$ In parallel, $\dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$, and then invert.
3. 3Find the total current from $V = IR$.
4. 4Work back to the branch or component the question asks about.
5. 5Find the power last with $P = VI$, then the energy with $E = Pt$, with $t$ in seconds.