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AP Physics 1- 3.5 Power- Study Notes- New Syllabus

AP Physics 1-3.5 Power – Study Notes

AP Physics 1-3.5 Power – Study Notes -AP Physics 1 – per latest Syllabus.

Key Concepts:

  • Power
  • Average Power
  • Instantaneous Power
  • Efficiency 

AP Physics 1-Concise Summary Notes- All Topics

Power

Power is the rate at which energy changes with respect to time, either by transfer into or out of a system or by conversion from one type to another within a system. (3.5.A.1)

Average Power:

The amount of energy transferred or converted divided by the time interval:

\( P_\text{avg} = \dfrac{\Delta E}{\Delta t} \)

Because work is the change in energy due to a force, average power can also be expressed as: 

\( P_\text{avg} = \dfrac{W}{\Delta t} \)

Instantaneous Power:

The power delivered to an object by the component of a constant force parallel to the object’s velocity:

\( P = \vec{F} \cdot \vec{v} = F v \cos\theta \), where \( \theta \) is the angle between force and velocity vectors.

Power is a scalar quantity, even though it depends on vector quantities like force and velocity.

  • The SI unit of power is the Watt (W): \( 1 \, \text{W} = 1 \, \text{J/s} \).

High power implies rapid energy transfer or conversion; low power implies slow transfer.

  • For systems with variable forces or velocity, instantaneous power is more useful than average power.

Power can be positive or negative:

    • Positive power: Force and velocity components in the same direction → energy added to system.
    • Negative power: Force and velocity components in opposite directions → energy removed from system (e.g., braking).

In mechanical systems, the efficiency can be expressed in terms of power:

\( \text{Efficiency} = \dfrac{\text{useful power output}}{\text{total power input}} \times 100\% \)

Example :

A \( 10 \, \text{kg} \) weight is lifted vertically \( 2 \, \text{m} \) in \( 4 \, \text{s} \). Calculate the average power delivered.

▶️Answer/Explanation

Step 1: Energy change = work done against gravity → \( \Delta E = m g h = 10 \cdot 9.8 \cdot 2 = 196 \, \text{J} \).

Step 2: Average power → \( P_\text{avg} = \dfrac{\Delta E}{\Delta t} = \dfrac{196}{4} = 49 \, \text{W} \).

Final Answer: Average power = 49 W.

Example : 

A car is moving with velocity \( \vec{v} = 20 \, \text{m/s} \) and a constant force \( \vec{F} = 500 \, \text{N} \) acts along the direction of motion. Find the instantaneous power.

▶️Answer/Explanation

Step 1: Use instantaneous power formula → \( P = \vec{F} \cdot \vec{v} = F v \cos\theta \).

Step 2: Force is parallel to velocity → \( \theta = 0^\circ \) → \( \cos 0^\circ = 1 \).

Step 3: Substitute values → \( P = 500 \cdot 20 \cdot 1 = 10{,}000 \, \text{W} \).

Final Answer: Instantaneous power = 10 kW.

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