Work, Energy and Power · Lesson 11 of 16
Power and Time-Dependent Energy Transfer
“Distinguish work from its rate and derive motion under constant force and constant power.”
• Calculate average and instantaneous power. • Derive P = F·v. • Compare constant-force and constant-power motion.
Same work, different rates
Two people may each lift the same load through the same height. If one takes half as long, that person delivers twice the average useful power. Work measures energy transferred; power measures how quickly it is transferred. A force can do a large amount of work slowly, and a brief process can have high power while transferring comparatively little total energy.
Power of an individual force
The force component along velocity adds or removes energy. Perpendicular force gives zero power. A braking force gives negative power on a moving vehicle. Net power equals dK/dt for a particle. Motor power can exceed net kinetic-energy power when some power lifts a load or compensates resistance. At zero instantaneous speed, a finite force has zero instantaneous mechanical power even if it can start accelerating the body.
Units and energy consumption
| Quantity | Unit |
|---|---|
| Power | 1 W = 1 J/s |
| Power dimensions | [M L² T⁻³] |
| 1 kilowatt | 1000 W |
| 1 horsepower | 746 W, approximately |
| 1 kilowatt-hour | 3.6×10⁶ J of energy |
Problem
A 100 W bulb operates for 10 h. How much energy does it use?
- 1.Energy = power×time = 100 W×10 h = 1000 Wh = 1 kWh.
- 2.In joules this is 100×36000 = 3.6×10⁶ J.
- 3.The bulb’s power remains 100 W; the 1 kWh is the accumulated energy, not a new power rating.
Constant acceleration from rest
For a fixed mass with constant net force F, acceleration a = F/m is constant. Starting from rest gives v = at. The net instantaneous power is therefore Fv = ma²t. It grows linearly with time because the same force acts through a greater displacement each second as the body speeds up.
Problem
A 2 kg mass accelerates from rest at 3 m/s² for 4 s under the net force. Find final and average net power.
- 1.Force is 6 N and final velocity is 12 m/s. Final power is 6×12 = 72 W.
- 2.The kinetic-energy gain is ½×2×12² = 144 J. Average power is 144/4 = 36 W.
- 3.Because power rises linearly from zero to 72 W, the average happens to be half its final value in this case.
Constant power gives a different acceleration
For the formal start-from-rest case, v is proportional to t^(1/2). Integrating velocity gives x − x0 = (2/3)√(2P/m)t^(3/2), so displacement is proportional to t^(3/2), not t². The model requires an infinite force at exactly zero speed because F = P/v. Real motors have a force or torque limit and cannot deliver ideal constant mechanical power from the exact rest instant. The scaling is an idealisation or applies after a finite-speed start.
Check your understanding
Quiz
A motor transfers 600 J in 3 s. Average power is
Explanation: Pav=W/Δt.
A force perpendicular to velocity has instantaneous power
Explanation: The dot product is zero.
Constant net force from rest gives P proportional to
Explanation: F is constant while v=at.
Ideal constant net power from rest gives x proportional to
Explanation: v is proportional to √t and integration adds one to the exponent.
Why is ideal constant positive power problematic at rest?
Explanation: A real actuator has finite force or torque.
Key Takeaways
• Power is the rate of energy transfer. • P = F·v applies instantaneously to a specified force. • Constant force and constant power lead to different motion laws.