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IB MYP 4-5 Physics- Power  & Efficiency – Study Notes

IB MYP 4-5 Physics- Power  & Efficiency - Study Notes - New Syllabus

IB MYP 4-5 Physics-Power  & Efficiency – Study Notes

Key Concepts

  • Power  & Efficiency 

IB MYP 4-5 Physics Study Notes – All topics

Power

Power

Power is the rate at which work is done or the rate at which energy is transferred or transformed.

The formula for power is:

P = \(\dfrac{W}{t}\) or \(P = \dfrac{E}{t}\)

Where:

  • \(P\) = Power (watts, W)
  • \(W\) = Work done (joules, J)
  • \(E\) = Energy transferred (joules, J)
  • \(t\) = Time taken (seconds, s)

In terms of force and velocity, power can also be expressed as:

\(P = F \cdot v\)

Where:

  • \(F\) = Force applied (newtons, N)
  • \(v\) = Velocity (m/s)

Key Notes

  • Measured in watts (W), where 1 W = 1 J/s.
  • Higher power means more work done in less time.
  • Power is scalar; it does not have direction.

Example:

A crane lifts a 2000 N load to a height of 15 m in 30 seconds. Calculate the power output.

▶️ Answer/Explanation

Work done: \(W = F \cdot d = 2000 \times 15 = 30000 \ \text{J}\)

Power: \(P = \dfrac{30000}{30} = 1000 \ \text{W}\)

Final Answer: \(\boxed{1000 \ \text{W}}\)

Example:

A car engine exerts a constant force of 500 N while moving at 20 m/s. Find its power output.

▶️ Answer/Explanation

Power: \(P = F \cdot v = 500 \times 20 = 10000 \ \text{W}\)

Final Answer: \(\boxed{10.0 \ \text{kW}}\)

Efficiency

Efficiency

Efficiency is the measure of how effectively input energy is converted into useful output energy or work. It is expressed as a ratio or percentage.

 

\(\text{Efficiency} = \dfrac{\text{Useful Output Energy (or Power)}}{\text{Total Input Energy (or Power)}} \times 100\%\)

Where:

  • Efficiency is always less than or equal to 100%.
  • Friction, heat loss, and sound loss reduce efficiency.

Formula (Energy-based):

\(\eta = \dfrac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%\)

Formula (Power-based):

\(\eta = \dfrac{\text{Useful power output}}{\text{Total power input}} \times 100\%\)

Formula (Work-based):

\(\eta = \dfrac{\text{Work output}}{\text{Work input}} \times 100\%\)

Formula (For mechanical advantage systems):

\(\eta = \dfrac{\text{Mechanical Advantage}}{\text{Velocity Ratio}} \times 100\%\)

Key Notes:

  • 100% efficiency means no energy is wasted — practically impossible in real systems.
  • Efficiency can be increased by reducing energy losses (e.g., using lubrication, better insulation).
  • Often calculated for machines, engines, and power plants.

Example:

A motor consumes 2000 J of electrical energy and produces 1500 J of useful mechanical work. Calculate its efficiency.

▶️ Answer/Explanation

Efficiency = \(\dfrac{1500}{2000} \times 100\% = 75\%\)

Final Answer: \(\boxed{75\%}\)

Example:

An engine outputs 300 kW of power for a fuel input of 400 kW. Find the efficiency.

▶️ Answer/Explanation

Efficiency = \(\dfrac{300}{400} \times 100\% = 75\%\)

Final Answer: \(\boxed{75\%}\)

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