Power

Section: Work, Energy and Power | Syllabus: Cambridge AS Level Physics (9702)

Defining Power

Power is the rate of transfer of energy or the rate at which work is done.

P = (W/t)

Units: Watts (W). 1W = 1 J s^-1.

Worked Example: Race up Stairs

Question: A 60 kg runner races up a tall building (vertical height 390 m) in 17 minutes 22 seconds.

(a) Calculate the work done.

(b) Calculate the average power developed.

Answer

(a) Work Done: Lifting mass against gravity.

W = mg Δ h = 60 × 9.81 × 390 = 230,000 J (230 kJ).

(b) Power: Convert time to seconds.

t = (17 × 60) + 22 = 1042 s.

P = (2.3 × 10^5/1042) ≈ 220 W

To keep a vehicle moving at constant speed against resistive forces, the engine must deliver power.

P = Fv

Where F is the driving force (equal to resistive forces at constant speed) and v is velocity.

Worked Example: Car on Highway

Question: A car of mass 1400 kg travelling at a constant velocity of 27 m s^-1 has total resistive forces of 820 N.

(a) Calculate the power developed.

(b) If the car accelerates from 27 m s^-1 with a driving force of 1100 N, calculate the initial acceleration.

Answer

(a) Power:

P = Fv = 820 × 27 = 22,140 W = 22 kW

(b) Acceleration:

Resultant Force = Driving Force - Resistive Force = 1100 - 820 = 280 N.

a = (F/m) = (280/1400) = 0.20 m s^-2

Constant Speed: When a question states "constant speed", it implies the driving force equals the resistive force (F_driving = F_resistive). Use this F in P=Fv.

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