Efficiency & Power (Edexcel GCSE Physics): Revision Note

An electric motor has an efficiency of 35%. It lifts a 7.2 kg load through a height of 5 m in 3 s.

Calculate the power of the motor.

 Answer:

Step 1: List the known quantities

• Efficiency = 35%

• Mass, m = 7.2 kg

• Height, h = 5 m

• Time, t = 3 s

• Gravitational field strength, g = 10 N/kg 

Step 2: Write down the efficiency equation 

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{power}\mathrm{output}}{\mathrm{total}\mathrm{power}\mathrm{input}}\times 100\%$ 

Step 3: Rearrange to make power input the subject

$\mathrm{useful}\mathrm{power}\mathrm{input}=\frac{\mathrm{power}\mathrm{output}\times 100}{\mathrm{efficiency}}$  

Step 4: Calculate the power output

$P=\frac{E}{t}$

• Energy was transferred from the kinetic store of the motor to the gravitational potential store of the load

$∆GPE=m\times g∆h$

$∆GPE=7.2\times 10\times 5$

$∆GPE=360\mathrm{J}$ 

• Therefore, the power output is

$P=\frac{360}{3}$

$P=120\mathrm{W}$ 

Step 5: Substitute values into power input equation

$\mathrm{useful}\mathrm{power}\mathrm{input}=\frac{\mathrm{power}\mathrm{output}\times 100}{\mathrm{efficiency}}$

$\mathrm{useful}\mathrm{power}\mathrm{input}=\frac{120\times 100}{35}$

$\mathrm{useful}\mathrm{power}\mathrm{input}=343\mathrm{J}$

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

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 (just the % sign)