Use total work divided by elapsed time without assuming an instantaneous force is constant.
Syllabus 4.5
Notes and key methods
Average power is total work done divided by elapsed time. A lifting device that raises a mass through vertical height h from rest to rest, with no resistance, does work Mgh because there is no net change in kinetic energy.
An instantaneous power need not equal the interval average. For a device supplying constant power P over time t, transferred energy is Pt. Keep joules, seconds and watts consistent; multiplying by time and dividing by time answer different questions.
Check the conditions
Choose and retain a positive direction; distinguish signed quantities from magnitudes.
Use SI units and \(\displaystyle g= \frac{10m}{s^{2}}\) wherever gravity is relevant.
Apply only the model and time interval stated in the question.
Three worked examples
Example 1 · Average power and work over an interval
A device raises a 5 kg load through 10 m vertically in 3 s, starting and ending at rest. Neglect resistance and use \(\displaystyle g= \frac{10m}{s^{2}}\). Separately, a constant-power device supplies 23 W.
Find the work done by the lifting device in J.
Find the average power of the lifting device in W.
Find the energy transferred by the separate constant-power device in 5 s, in J.
How long does the separate constant-power device take to transfer 92 J? Give seconds.
Show the reasoning
(a)
Account for the gravitational potential-energy rise and the equal initial/final kinetic energies.
The load has no net kinetic-energy change, so lifting work equals Mgh=500 J.
Answer: \(500\).
(b)
Use total lifting work divided by the specified elapsed time.
Work divided by total time gives \(\displaystyle \frac{500}{3}\) W.
Answer: \(\displaystyle \frac{500}{3}\).
(c)
Multiply watts by seconds to recover transferred energy.
Constant power times duration gives 115 J.
Answer: \(115\).
(d)
Rearrange E=Pt for elapsed time using the separate device’s power.
Energy divided by its constant power gives 4 s.
Answer: \(4\).
Example 2 · Average power and work over an interval
A device raises a 6 kg load through 11 m vertically in 4 s, starting and ending at rest. Neglect resistance and use \(\displaystyle g= \frac{10m}{s^{2}}\). Separately, a constant-power device supplies 24 W.
Find the work done by the lifting device in J.
Find the average power of the lifting device in W.
Find the energy transferred by the separate constant-power device in 6 s, in J.
How long does the separate constant-power device take to transfer 120 J? Give seconds.
Show the reasoning
(a)
Account for the gravitational potential-energy rise and the equal initial/final kinetic energies.
The load has no net kinetic-energy change, so lifting work equals Mgh=660 J.
Answer: \(660\).
(b)
Use total lifting work divided by the specified elapsed time.
Work divided by total time gives \(165\) W.
Answer: \(165\).
(c)
Multiply watts by seconds to recover transferred energy.
Constant power times duration gives 144 J.
Answer: \(144\).
(d)
Rearrange E=Pt for elapsed time using the separate device’s power.
Energy divided by its constant power gives 5 s.
Answer: \(5\).
Example 3 · Average power and work over an interval
A device raises a 7 kg load through 4 m vertically in 5 s, starting and ending at rest. Neglect resistance and use \(\displaystyle g= \frac{10m}{s^{2}}\). Separately, a constant-power device supplies 25 W.
Find the work done by the lifting device in J.
Find the average power of the lifting device in W.
Find the energy transferred by the separate constant-power device in 3 s, in J.
How long does the separate constant-power device take to transfer 150 J? Give seconds.
Show the reasoning
(a)
Account for the gravitational potential-energy rise and the equal initial/final kinetic energies.
The load has no net kinetic-energy change, so lifting work equals Mgh=280 J.
Answer: \(280\).
(b)
Use total lifting work divided by the specified elapsed time.
Work divided by total time gives \(56\) W.
Answer: \(56\).
(c)
Multiply watts by seconds to recover transferred energy.
Constant power times duration gives 75 J.
Answer: \(75\).
(d)
Rearrange E=Pt for elapsed time using the separate device’s power.
Energy divided by its constant power gives 6 s.
Answer: \(6\).
Common mistakes
Average power uses the complete work and complete interval.
Do not substitute a starting zero speed into \(\displaystyle \frac{P}{v}\) for a whole lifting interval.