Lesson Explainer: Power as the Rate of Work Mathematics • Third Year of Secondary School

In this explainer, we will learn how to define the power of a force as the derivative of the work done by the force.

If the net force on a body does work on it, the work done by the force on the body is equal to the change in the kinetic energy of the body.

Kinetic energy transferred to a body due to work done by a net force may then be transferred to gravitational potential energy and may also be dissipated by resistive forces acting on the body. For clarity, we limit this explainer to situations involving forces changing the kinetic energy of bodies without further energy transfers occurring.

Let us define the work done on a body by a force.

Let us define kinetic energy.

When a force does work on a body to change its kinetic energy, the kinetic energy of the body changes over the time interval in which the force acts. The force does work on the body at a certain rate. The rate of change of work done on the body by the force is the power supplied by the force.

A constant net force acting on a body produces an acceleration of the body, , in the direction of . For a body that is at rest at a point before the force acts on it, the displacement of the body from in the direction of is given by where is the displacement of the body during a time interval, which is zero at .

The value of can be substituted for the displacement of the body in the direction of the force to give where is the work done between and the instant .

The change of with from to is shown in the following graph. The magnitudes of and are both 1.

The rate at which work done between and the instant increases is equal to the gradient of the curve at the instant . The gradient of the line is the derivative with respect to of which can be written as the time derivative of the equation as follows:

As the rate at which work done by a force on a body is the power supplied by the force, this expression is the instantaneous power supplied to the body by the constant force. The instantaneous power is a function of .

If a force is not constant, it does not produce a constant acceleration, so the instantaneous power, , supplied by a force can be given in a more generalizable form by where is the velocity of the body.

It might initially seem that a constant force acting on a body should supply a constant power to the body. This impression can be seen to be incorrect, however, by comparing an increase in velocity to the increase in kinetic energy due to the increase in velocity.

Consider a body that accelerates uniformly from rest. In an interval , there is an increase in the velocity, , of the body, where the initial velocity is zero:

The kinetic energy of the body increases by a factor , where

Now, consider a second increase in velocity that is equal to the first increase, also occurring in an interval :

As it follows that

An increase in velocity corresponds to an increase in kinetic energy by a factor , where

Therefore,

The increases in kinetic energy are unequal, and so the rate of work done on the body is unequal for two equal time intervals.

Suppose instead that, in equal successive time intervals, equal work is done on the body, so that

Again, assuming that the body is initially at rest, at , the velocity of the body is and at , the velocity of the body is

For work to be done on the body at constant rate, which cannot be the case for nonzero positive , and for an object accelerated by a constant net force, would have to be both nonzero and positive. If is positive and nonzero, it must be the case that

So, if in equal successive time intervals equal work is done on a body of a fixed mass by a force, the acceleration of the body and, hence, of the force acting on it must decrease for each successive time interval. A constant power means that the force acting to supply the power is a function of time.

As well as the instantaneous power supplied by a force, the average power supplied by a force can be determined. The average power supplied to a body over a time interval is equal to the work done during , divided by .

On a work–time graph, the average power is therefore equal to the gradient of the straight line that intersects the value of at the initial and final time values of an interval, as shown in the following graph by the dashed line, where is the interval to .

A constant force does work on a body as a function of time that the force acts for. If the force is not constant, however, but is a function of time, the rate at which the force does work on a body can be constant. Such a force is said to supply a constant power.

If a force supplies an instantaneous power that is the same at all instants in which the force acts, the average power supplied by the force is equal to the instantaneous power of the force at each instant. Therefore, a work–time graph will have a constant gradient for a force that supplies a constant power, just as the average power of a force is given by the gradient of a straight line.

Let us look at an example where a constant value of power is used.

Now let us look at an example where instantaneous power is used.

Example 2: Calculating Power Based on Work and Time

The power of a machine is given by the relation , where is the time elapsed in seconds. Find the work done by the machine in the first 8 seconds.

Answer

The power supplied by the machine varies with time, according to the equation so the work done by the machine is the work done between and some instant .

The question asks for the work done by the machine in an 8-second time interval starting at . As power is the time rate of work, work is the product of power and time. The work done in the time interval corresponds to the area between the line and the axes of the following power–time graph.

The equation of the line is

The work can be determined by integrating with respect to :

As an alternative to using integration, the area could be determined as follows.

The power at is given by

The area consists of that of a rectangle with side lengths of 8 and 7 added to that of a right triangle with a base length of 8 and a height of . The areas are found as follows: and the total area is

The units of power are not stated in the question, so the answer must be expressed as

In the previous example, it was possible to answer the question without the use of integration. Now, let us look at an example similar to the previous one for which integration is necessary.

Example 3: Calculating Power Based on Work and Time

The power of an engine at time seconds is given by . Find the work done by the engine between and .

Answer

The power supplied by the engine varies with time. As power is the time rate of work, work is the product of power and time. The work done in the time interval corresponds to the area between the curve, the -axis, and the dashed lines shown by the following power–time graph.

The area to be calculated is not that of a regular shape but can be determined by integration as follows:

Therefore, the work done by the engine between and is 96 J.