What is Power?
This topic is part of the HSC Physics course under the section Forces, Acceleration and Energy.
HSC Physics Syllabus
- conduct investigations over a range of mechanical processes to analyse qualitatively and quantitatively the concept of average power 𝑃=Δ𝐸Δ𝑡, 𝑃= 𝐹∥𝑣=𝐹𝑣cos𝜃 including but not limited to:
Introduction to Power
What is Power?
Power (P) is the rate at which work is done or energy is transferred over time. It is defined as the work done per unit time or the energy transferred per unit time. The formula to calculate power is:
$$P = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$$
where:
- W is the work done (measured in Joules, J)
- P is the power (measured in Watts, W or Joules per second, J/s)
- ∆E is the change in energy (measured in Joules, J)
- t is the change in time (measured in seconds, s)
Like energy and work, power is a scalar quantity.
Mechanical Power
Mechanical power is the rate of change in work done related to the motion or position of a mass.
In the case of mechanical energy (related to motion or position of a mass), work is given by the following equation.
$$W = Fs\cos{\theta}$$
$$\Delta E = Fs\cos{\theta}$$
Therefore, an equation for mechanical power in terms of force and velocity can be derived:
$$P = \frac{Fs\cos{\theta}}{\Delta t}$$
$$P = Fv\cos{\theta}$$
- P is the mechanical power (measured in W or J/s)
- F is the force applied to do work (measured in N)
- v is the average velocity throughout the displacement (measured in m/s)
- θ is the angle between the force vector and displacement of the mass
Work Done in Uniformly Accelerated Rectilinear Motion
Work done in motion with acceleration is transferred to an object's kinetic energy. Therefore, the power exerted in motion with accelerated equals to the change in kinetic energy divided by the change in time.
$$P = \frac{\Delta K}{\Delta t}$$
Alternatively, the power exerted can be determined by considering the magnitude of acceleration.
$$P = \frac{Fs\cos{\theta}}{\Delta t}$$
Applying Newton's second law:
$$P = \frac{mas\cos{\theta}}{\Delta t}$$
Calculating Power – Example 1
Consider a car of mass 1000 kg, which accelerates from rest to a speed of 20 m/s in 10 seconds over a straight distance of 100 meters.
Power can be determined by considering the rate of change in kinetic energy:
$$P = \frac{\Delta K}{\Delta t}$$
$$P = \frac{\frac{1}{2}mv^2-\frac{1}{2}mu^2}{\Delta t}$$
$$P = \frac{\frac{1}{2}(1000)(20)^2-\frac{1}{2}(1000)(0)^2}{10}$$
$$P = 20000 \hspace{2mm} J \hspace{1mm} s^{-1}$$
Alternatively, power can be determined by considering the acceleration and displacement:
$$a = \frac{20 - 0}{10}$$
$$a = 2 \hspace{2mm} m \hspace{1mm} s^{-1}$$
$$P = \frac{mas\cos{\theta}}{\Delta t}$$
$$P = \frac{(1000)(2)(100)\cos{0º}}{10}$$
$$P = 20000 \hspace{2mm} J \hspace{1mm} s^{-1}$$
Power and Work Done Against Gravity
When a mass is lifted from the ground, work is done to increase its gravitational potential energy. This requires the magnitude of applied force to be at least equal to the mass's weight force.
$$W = \Delta U$$
In this case, the power exerted equals the work done divided the time over which it is exerted:
$$P = \frac{\Delta U}{\Delta t}$$
$$P = \frac{mg\Delta h}{\Delta t}$$
Calculating Power – Example 2
Imagine a motor is used to lift a 50 kg mass to a height of 5 meters in 10 seconds. First, we need to calculate the work done, which is equal to the change in potential energy in this case:
$$W = mg \Delta h$$
$$W = 50 \times 9.8 \times 5$$
$$W = 2452.5 \hspace{2 mm} J$$
Now, we can calculate the power required by the motor using the formula:
$$P = \frac{W}{t}$$
$$P = \frac{2452.5}{10}$$
$$P = 245.25 \hspace{2mm} J \hspace{1mm} s^{-1}$$
The power needed by the motor to lift the weight is 245.25 Watts or Joules per second.
Power Exerted Against Friction and Air Resistance
Friction, air resistance and rolling friction are examples of non-conservative forces. When non-conservative forces are present, an object's mechanical energy (kinetic energy and potential energy) is not conserved because it is partially transformed into other forms such as heat.
Calculating Power – Example 3
Suppose a car with a constant driving force of 4000 N is moving at a constant speed of 20 m/s. The force is applied in the direction of the car's motion.
The power exerted by the car engine can be calculated as follows:
$$P = Fv\cos{\theta}$$
$$P = (4000)(20)\cos{0º}$$
$$P = 80000 \hspace{2mm} J \hspace{1mm} s^{-1}$$
In this case, 4000 N of force is necessary to do work against friction between the car's tyres and the road. 80000 J/s is also the power exerted against friction. This means the magnitude of kinetic friction also equals 4000 N, resulting in a net force of 0 N. When the net force is 0, the car is able to travel at a constant speed (Newton's second law, F = ma).