Lesson Explainer: Power as the Rate of Work Mathematics

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.

Definition: Work Done on a Body by a Force

The work done on a body by a force is dependent on the force that acts on the body and the displacement of the body, according to the formula ๐‘Š=โƒ‘๐นโ‹…โƒ‘๐‘ , where โƒ‘๐น is the force and โƒ‘๐‘  is the displacement of the body.

Let us define kinetic energy.

Definition: Kinetic Energy

The kinetic energy of a body is dependent on the mass and speed of the body, according to the formula ๐ธ=12๐‘š๐‘ฃ,K๏Šจ where ๐‘š is the mass of the body and ๐‘ฃ is the speed of the body.

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 โƒ‘๐น acting on a body produces an acceleration of the body, โƒ‘๐‘Ž, in the direction of โƒ‘๐น. For a body that is at rest at a point ๐‘ƒ before the force acts on it, the displacement of the body from ๐‘ƒ in the direction of โƒ‘๐‘Ž is given by โƒ‘๐‘ =12โƒ‘๐‘Ž๐‘ก,๏Šจ where โƒ‘๐‘  is the displacement of the body during a time interval, which is zero at ๐‘ก=0.

The value of โƒ‘๐‘  can be substituted for the displacement of the body in the direction of the force to give ๐‘Š=โƒ‘๐นโ‹…โƒ‘๐‘ =12โƒ‘๐นโ‹…โƒ‘๐‘Ž๐‘ก,๏Šจ where ๐‘Š is the work done between ๐‘ก=0 and the instant ๐‘ก.

The change of ๐‘Š with ๐‘ก from ๐‘ก=0 to ๐‘ก=5 is shown in the following graph. The magnitudes of โƒ‘๐น and โƒ‘๐‘Ž are both 1.

The rate at which work done between ๐‘ก=0 and the instant ๐‘ก increases is equal to the gradient of the curve at the instant ๐‘ก. The gradient of the line is the derivative with respect to ๐‘ก of ๐‘Š=12โƒ‘๐นโ‹…โƒ‘๐‘Ž๐‘ก,๏Šจ which can be written as the time derivative of the equation as follows: dd๐‘Š๐‘ก=โƒ‘๐นโ‹…โƒ‘๐‘Ž๐‘ก.

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 ๐‘ก.

If a force is not constant, it does not produce a constant acceleration, so the instantaneous power, ๐‘ƒ, supplied by a force can be given in a more generalizable form by ๐‘ƒ=โƒ‘๐นโ‹…โƒ‘๐‘ฃ, where โƒ‘๐‘ฃ 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 ฮ”๐‘ก, there is an increase in the velocity, ฮ”๐‘ฃ, of the body, where the initial velocity is zero: ฮ”๐‘ฃ=๐‘ฃโˆ’0=๐‘ฃ.๏Šง๏Šง

The kinetic energy of the body increases by a factor ฮ”๐ธ๏Œช, where ฮ”๐ธ=๐‘ฃ.๏Œช๏Šง๏Šจ

Now, consider a second increase in velocity that is equal to the first increase, also occurring in an interval ฮ”๐‘ก: ฮ”๐‘ฃ=๐‘ฃโˆ’๐‘ฃ=ฮ”๐‘ฃ.๏Šจ๏Šจ๏Šง

As ๐‘ฃโˆ’๐‘ฃ=๐‘ฃ,๏Šจ๏Šง๏Šง it follows that ๐‘ฃ=2๐‘ฃ.๏Šจ๏Šง

An increase in velocity ฮ”๐‘ฃ๏Šจ corresponds to an increase in kinetic energy by a factor ฮ”๐ธ๏Œช๏Šจ, where ฮ”๐ธ=๐‘ฃโˆ’๐‘ฃ=(2๐‘ฃ)โˆ’๐‘ฃ=4๐‘ฃโˆ’๐‘ฃ=3๐‘ฃ.๏Œช๏Šจ๏Šจ๏Šจ๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ

Therefore, ฮ”๐ธ=3ฮ”๐ธ.๏Œช๏Šจ๏Œช

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 ฮ”๐ธ=ฮ”๐ธ.๏Œช๏Œช๏Šจ

Again, assuming that the body is initially at rest, at ๐‘ก=ฮ”๐‘ก, the velocity of the body is 0+ฮ”๐‘ฃ=ฮ”๐‘ฃ, and at ๐‘ก=2ฮ”๐‘ก, the velocity of the body is ฮ”๐‘ฃ+ฮ”๐‘ฃ.๏Šจ

For work to be done on the body at constant rate, ฮ”๐‘ฃ=(ฮ”๐‘ฃ+ฮ”๐‘ฃ)โˆ’ฮ”๐‘ฃ,๏Šจ๏Šจ๏Šจ๏Šจ which cannot be the case for nonzero positive ฮ”๐‘ฃ, and for an object accelerated by a constant net force, ฮ”๐‘ฃ would have to be both nonzero and positive. If ฮ”๐‘ฃ is positive and nonzero, it must be the case that ฮ”๐‘ฃ=ฮ”๐‘ฃ๏€ปโˆš2โˆ’1๏‡.๏Šจ

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 ฮ”๐‘ก is equal to the work done during ฮ”๐‘ก, divided by ฮ”๐‘ก.

On a workโ€“time graph, the average power is therefore equal to the gradient of the straight line that intersects the value of ๐‘Š at the initial and final time values of an interval, as shown in the following graph by the dashed line, where ฮ”๐‘ก is the interval ๐‘ก=0 to ๐‘ก=5.

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 workโ€“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.

Example 1: Finding the Time Needed by a Car Moving with a Constant Given Power to Reach a Certain Velocity

Find the time taken for a car of 1โ€Žโ€‰โ€Ž236 kg to reach a speed of 126 km/h, given that the car started from rest and that the power of the engine is constant and equal to 103 metric horsepower.

Answer

The car is initially at rest, with no kinetic energy. When the car has a speed of 126 km/h, this is equivalent to a speed in metres per second of ๐‘ฃ=126ร—10003600=35/.ms

The kinetic energy of the car when it has a speed of 35 m/s is given by ๐ธ=12ร—1236ร—35=757050.๏Œช๏Šจjoules

A power of 1 metric horsepower is equal to 735 W. The power supplied by the engine is, therefore, given by ๐‘ƒ=103ร—735=75705.W

The length of the time interval in which the car accelerates is the kinetic energy increase of the car divided by the power supplied by the engine, so ฮ”๐‘ก is given by ฮ”๐‘ก=75705075705=10.s

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 ๐‘ƒ=3๐‘ก+7, where ๐‘ก 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 ๐‘ƒ=3๐‘ก+7, so the work done by the machine is the work done between ๐‘ก=0 and some instant ๐‘ก.

The question asks for the work done by the machine in an 8-second time interval starting at ๐‘ก=0. 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 powerโ€“time graph.

The equation of the line is ๐‘ƒ=3๐‘ก+7,0โ‰ค๐‘กโ‰ค8.

The work can be determined by integrating ๐‘ƒ(๐‘ก) with respect to ๐‘ก: ๐‘Š=๏„ธ3๐‘ก+7๐‘ก=๏”32๐‘ก+7๐‘ก๏ ,๐‘Š=32(64โˆ’0)+56=152.๏Šฎ๏Šฆ๏Šจ๏Šฎ๏Šฆd

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

The power at ๐‘ก=8 is given by ๐‘ƒ=3(8)+7=31.๏Šฎ

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 31โ€“7. The areas are found as follows: ๐ด=8ร—7=56,๐ด=8ร—31โˆ’72=96,rectangletriangle and the total area is 56+96=152.

The units of power are not stated in the question, so the answer must be expressed as ๐‘Š=152.workunits

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 ๐‘ก seconds is given by ๐‘ƒ(๐‘ก)=๏€น12๐‘ก+8๐‘ก๏…๏ŠจW. Find the work done by the engine between ๐‘ก=2s and ๐‘ก=3s.

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 ๐‘ฅ-axis, and the dashed lines shown by the following powerโ€“time graph.

The area to be calculated is not that of a regular shape but can be determined by integration as follows: ๐‘Š=๏„ธ12๐‘ก+8๐‘ก๐‘ก=๏”123๐‘ก+82๐‘ก๏ =4๏‘๐‘ก+๐‘ก๏๐‘Š=4๏€น๏€น3+3๏…โˆ’๏€น2+2๏…๏…๐‘Š=4(36โˆ’12)=96.๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจd

Therefore, the work done by the engine between ๐‘ก=2s and ๐‘ก=3s is 96 J.

Key Points

  • The power supplied by a force is the rate at which a force does work on a body and is the time derivative of the work done on the body by the force.
  • The value of the power supplied by a force can be either an instantaneous or an average value.
  • When a body is accelerated by a constant net force, the rate at which work is done on the body by the force is directly proportional to the time for which the force has acted on the body.

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