AP Physics C Mechanics- 3.5 Power- Study Notes- New Syllabus
AP Physics C Mechanics- 3.5 Power – Study Notes
AP Physics C Mechanics- 3.5 Power – Study Notes – per latest Syllabus.
Key Concepts:
- Power
- Average Power,
- Instantaneous Power
- Power from Constant Force (parallel component)
Power
Power is the rate at which energy changes with respect to time — either by being transferred into or out of a system, or converted from one form to another within the system. It describes how quickly work is done or energy is transformed.
Concept of Power and Energy Change
Power quantifies how fast energy transfer or conversion occurs. It can represent:
- Work done per unit time by a force, or
- The rate at which potential or kinetic energy changes.
\( \mathrm{P = \dfrac{dE}{dt}} \)
- \( \mathrm{P} \): power (watts, W = J/s)
- \( \mathrm{E} \): energy (J)
- \( \mathrm{t} \): time (s)
Average Power
The average power over a time interval is the total amount of energy transferred or work done divided by the time taken:
\( \mathrm{P_{avg} = \dfrac{W}{\Delta t} = \dfrac{\Delta E}{\Delta t}} \)
- \( \mathrm{W} \): total work done on or by the system
- \( \mathrm{\Delta t} \): time interval during which the work occurs
- It represents the average rate at which work is performed or energy is changed.
- Power is positive when energy is added to the system and negative when energy leaves the system.
Example
A motor lifts a \( \mathrm{50\,kg} \) crate vertically upward through a height of \( \mathrm{10\,m} \) in \( \mathrm{5.0\,s} \). Calculate the average power delivered by the motor. Take \( \mathrm{g = 9.8\,m/s^2} \).
▶️ Answer / Explanation
Step 1: Work done = increase in gravitational potential energy:
\( \mathrm{W = mgh = (50)(9.8)(10) = 4900\,J} \)
Step 2: Average power:
\( \mathrm{P_{avg} = \dfrac{W}{t} = \dfrac{4900}{5.0} = 980\,W} \)
Result: The motor’s average power output is \( \mathrm{980\,W} \).
Instantaneous Power
The instantaneous power is the rate at which work is being done or energy is being transferred at a specific instant in time:
\( \mathrm{P = \dfrac{dW}{dt}} \)
- From \( \mathrm{W = \vec{F} \cdot \vec{r}} \), differentiating gives:
\( \mathrm{P = \vec{F} \cdot \dfrac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}} \)
- \( \mathrm{\vec{F}} \): instantaneous force acting on the object
- \( \mathrm{\vec{v}} \): instantaneous velocity of the object
- The dot product ensures only the component of force in the direction of motion contributes to power.
Example
A car engine exerts a force of \( \mathrm{500\,N} \) on the car while it is moving at \( \mathrm{20\,m/s} \). Determine the instantaneous power output of the engine.
▶️ Answer / Explanation
Step 1: Use the formula for instantaneous power:
\( \mathrm{P = \vec{F} \cdot \vec{v}} \)
Step 2: Since force and velocity are in the same direction:
\( \mathrm{P = Fv = (500)(20) = 10{,}000\,W = 10\,kW} \)
Result: The engine delivers an instantaneous power of \( \mathrm{10\,kW} \).
Power from the Parallel Component of a Constant Force
When a constant force acts at an angle \( \mathrm{\theta} \) relative to an object’s velocity, only the component of the force that is parallel to the velocity contributes to the work and hence to power:
\( \mathrm{P = F_{\parallel} v = Fv \cos\theta} \)
- \( \mathrm{F_{\parallel} = F \cos\theta} \): parallel component of the force.
- \( \mathrm{v} \): magnitude of the object’s velocity.
- \( \mathrm{P} \) is positive if force and velocity point in the same direction (force adds energy), and negative if opposite (force removes energy).
Example
A person pulls a sled with a constant force of \( \mathrm{60\,N} \) at an angle of \( \mathrm{30°} \) above the horizontal. The sled moves at a constant velocity of \( \mathrm{2.0\,m/s} \). Find the instantaneous power delivered by the pulling force.
▶️ Answer / Explanation
Step 1: Use the power formula for constant force:
\( \mathrm{P = Fv\cos\theta} \)
Step 2: Substitute known values:
\( \mathrm{P = (60)(2.0)\cos(30°) = 120(0.866) = 104\,W} \)
Result: The instantaneous power output by the person is \( \mathrm{104\,W} \).