Efficiency (Cambridge (CIE) IGCSE Physics): Revision Note

An electric motor lifts a 7.2 kg load through a height of 5.0 m in 3 seconds. The efficiency of the motor is 35%.

Calculate the power supplied to the motor.

Answer:

Step 1: List the known quantities

• Mass of load, $m$ = 7.2 kg

• Change in height, $h$ = 5.0 m

• Acceleration of free fall, $g$ = 9.8 m/s²

• Time, $t$ = 3 s

• Efficiency = 35% or 0.35

Step 2: Calculate the power output of the motor

• The power supplied by the motor is the work done, or energy transferred, to lift the load

$\mathrm{power}\mathrm{output}=\frac{∆E}{t}$

• $∆E$ is equal to the change in gravitational potential energy as the load is lifted

$∆E_P=mg∆h$

$∆E_P=7.2\times 9.8\times 5.0=352.8\mathrm{J}$

• Therefore:

$\mathrm{power}\mathrm{output}=\frac{352.8}{3}=117.6\mathrm{W}$

Step 3: Write down the efficiency equation in terms of power

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{power}\mathrm{output}}{\mathrm{total}\mathrm{power}\mathrm{input}}$

Step 4: Rearrange to make power input the subject

$\mathrm{power}\mathrm{input}=\frac{\mathrm{power}\mathrm{output}}{\mathrm{efficiency}}$

Step 5: Substitute the values into the power input equation 

$\mathrm{power}\mathrm{input}=\frac{117.6}{0.35}$

$\mathrm{power}\mathrm{input}=336\mathrm{W}$

Efficiency can be given in a ratio (between 0 and 1) or percentage format (between 0 and 100 %)

If the question asks for efficiency as a ratio, give your answer as a fraction or decimal.

If the answer is required as a percentage, remember to multiply the ratio by 100 to convert it:

• if the ratio = 0.25, percentage = 0.25 × 100 = 25 %

Remember that efficiency has no units