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3.3Power and efficiency
When a machine performs work, it is important to know not only how much work is being done but also how much work is performed within a given time interval. A cyclist will perform a lot of work in a lifetime of cycling, but the same work can be performed by a powerful car engine in a much shorter time. Power is the rate at which work is being performed or the rate at which energy is being transferred.
When a quantity of work ÎW is performed within a time interval Ît the power developed is given by the ratio
P=ÎWÎt
Its unit is joules per second, and this is given the name watt (W): 1 W = 1 J sâ1.
Consider a constant force F, which acts on a body of mass m. The force does an amount of work F Îx in moving the body a small distance Îx along its direction. If this work is performed in time Ît, then
P=FÎxÎt
P=Fv
where v is the instantaneous speed of the body. This is the power produced in making the body move at speed v. As the speed increases, the power necessarily increases as well.
Consider an aircraft moving at constant speed on a straight-line path. If the power produced by its engines is P, and the force pushing it forward is F, then P, F and v are related by the equation above. But since the plane moves with no acceleration, the total force of air resistance must equal F. Hence the force of air resistance can be found simply from the power of the planeâs engines and the constant speed with which it coasts.
EXAM TIP
In many questions you may be asked to find the average power developed. It can be proven that this is given by FÂ¯Ã(v+u)2, where FÂ¯ is the average force and u and v the initial and final speeds.
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WORKED EXAMPLE 3.17
Estimate the minimum power required to lift a mass of 50.0 kg up a vertical distance of 12 m in 5.0 s.
Answer
The work done in lifting the mass is mgh:
W=mgh=50.0Ã9.81Ã12
W=5.89Ã103Â J
The power is therefore:
P=WÎt
P=5.89Ã1035.0â1200Â W
This is the minimum power required. In practice, the mass has to be accelerated from rest, which will require additional work and hence more power. There will also be frictional forces to overcome adding to the power actually needed.
CHECK YOURSELF 4
A boy of mass 60 kg jumps straight up into the air from a crouching position. His centre of mass is raised by 0.50 m when he is about to lose contact with the ground. After losing contact with the ground his centre of mass is lifted an additional 0.80 m. What is the average power developed? (Assume the boy pushes against the ground with a constant force and take g = 10 m sâ2.)
Efficiency
If a machine, such as an electric motor, is used to raise a load, electrical energy must be provided to the motor. This is the input energy to the motor. The motor uses some of this energy to do the useful work of raising the load. But some of the input energy is used to overcome frictional forces and therefore gets transferred to thermal energy. So the ratio:
Î·=Â useful energy outÂ Â actual energy inÂ  or Â useful power outÂ Â actual power inÂ 
is less than one. We call this ratio the efficiency, Î· of the machine.
Suppose that a body is being pulled up along a rough inclined plane with constant speed. The mass is 15 kg and the angle of the incline is 45Â°. There is a constant frictional force of 42 N opposing the motion.
The forces on the body are shown in Figure 3.32. Since the body has no acceleration, we know that
R=mgcosÎ¸=104.1Â N
F=mgsinÎ¸+f=104.1+42=146.1Â N
Let the force raise the mass a distance of 25 m along the plane. The work done by the force F is
W=146.1Ã25
W=3653Â Jâ3.7Ã103Â J
The force effectively raised the 15 kg a vertical height of 17.7 m (that is, 25 Ã sin 45Â°) and so increased the potential energy of the mass by mgh = 2605 J. The efficiency with which the force raised the mass is thus
efficiencyÂ =26053653=0.71

Figure 3.32: Forces on a body on an inclined plane: pulling force F, frictional force f, normal R and weight mg.

WORKED EXAMPLE 3.18
A 0.50 kg battery-operated toy train moves with constant velocity 0.30 m sâ1 along a level track. The power of the motor in the train is 2.0 W, and the total force opposing the motion of the train is 5.0 N.
aDetermine the efficiency of the trainâs motor.
bAssuming the efficiency and the opposing force stay the same, calculate the speed of the train as it climbs an incline of 10.0Â° to the horizontal.
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Answer
aThe power delivered by the motor is 2.0 W. Since the speed is constant, the force developed by the motor is 5.0 N.
The power used in moving the train is Fv = 5.0 Ã 0.30 = 1.5 W.
Hence the efficiency is
total power outÂ Â total power inÂ =1.5Â W2.0Â W=0.75
The efficiency of the trainâs motor is 0.75 (or 75%).
bThe component of the trainâs weight acting down the plane is mg sin Î¸ and the force opposing motion is 5.0 N. Since there is no acceleration (constant velocity), the net force F pushing the train up the incline is
F=mgsinÎ¸+5.0
F=0.50Ã9.81Ãsin10Â°+5.0
F=5.852Â Nâ5.9Â N
Thus
efficiencyÂ =5.852Ãv2.0
But from part a the efficiency is 0.75, so
â0.75=5.852Ãv2.0v=2.0Ã0.755.852v=0.26Â mÂ sâ1
TEST YOUR UNDERSTANDING
27The engine of a car is developing a power of 90 kW when it is moving on a horizontal road at a constant speed of 100 km hâ1.
aEstimate the total horizontal force opposing the motion of the car.
bAssuming the opposing force stays the same, what additional power must the engine deliver for the car to continue at the same speed up an incline making an angle of 5.0Â° to the horizontal? The mass of the car is 1200 kg.
28The motor of an elevator develops power at a rate of 2500 W.
aCalculate the speed that a 1200 kg load is being raised at.
bIn practice it is found that the load is lifted more slowly than indicated by your answer to a. Suggest reasons why this is so.
29A load of 50 kg is raised a vertical distance of 15 m in 125 s by a motor.
aEstimate the power necessary for this.
bThe power supplied by the motor is 80 W. Calculate the efficiency of the motor.
cThe same motor is now used to raise a load of 100 kg the same distance. The efficiency remains the same. Estimate how long this would take.
30The top speed of a racing car of mass 1200 kg whose engine is delivering 250 kW of power is 240 km hâ1. The car now moves on an uphill that makes an angle of 12Â° to the horizontal. Assuming the resistance force stays the same, what would the top speed of the car be?
31A 30 kg child and a 60 kg adult both run up the same flight of stairs in the same time. Who develops more power?
