Lesson Video: Power as the Rate of Work Mathematics

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

13:53

Video Transcript

In this video, we will learn how to define the power of a force as the derivative of the work done by the force. If a 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. In this video, we will not consider any further energy transfers. We will begin by defining the 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 such that π‘Š is equal to the dot or scalar product of 𝐅 and 𝐬, where 𝐅 and 𝐬 are the force and displacement vectors, respectively.

We recall that the kinetic energy of a body is dependent on the mass and speed such that the kinetic energy is equal to a half π‘šπ‘£ squared. 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. This rate of change of work done on the body by the force is the power supplied by the force. The power 𝑃 is therefore equal to dπ‘Š by d𝑑. We differentiate the work done with respect to time. Let’s now consider how this works in practice.

In the diagram shown, we have a constant net force vector 𝐅 acting on a body producing an acceleration vector 𝐚. Note that the acceleration and force act in the same direction. If the body is at rest at a point 𝑝 before the force acts on it, the displacement of the body is given by vector 𝐬 is equal to a half multiplied by vector 𝐚 multiplied by 𝑑 squared, where 𝐬 equals zero when 𝑑 equals zero. We recall that the work done is equal to the dot product of the force and displacement vectors. Replacing 𝐬 with a half πšπ‘‘ squared, the work done can be rewritten as the dot product of vector 𝐅 and a half πšπ‘‘ squared. The change of π‘Š at 𝑑 from 𝑑 equals zero to 𝑑 equals five is shown in the graph, where the magnitudes of vector 𝐅 and vector 𝐚 are both one.

The rate at which work done increases between 𝑑 equals zero and the instant 𝑑 is equal to the gradient of the curve at the instant 𝑑. Recalling the power is equal to dπ‘Š by d𝑑, and differentiating with respect to 𝑑, we have dπ‘Š by d𝑑 is equal to the dot product of 𝐅 and πšπ‘‘. If the force is not constant, it does not produce a constant acceleration. So the instantaneous power 𝑃 can be given by the more general form the dot product of 𝐅 and 𝐯, where vector 𝐯 is the velocity of the body. It is worth noting at this time that a constant force does not result in a constant power.

As well as finding out the instantaneous power supplied by a force, we can also calculate the average power. The average power is equal to the work done divided by the change in time, which is denoted by Δ𝑑. On our work–time graph, the average power is therefore equal to the gradient of the straight line that intersects the value of π‘Š at the initial and final time values of the interval in the graph shown from 𝑑 equals zero to 𝑑 equals five. Before looking at some specific questions, let’s now recall the formulae we will need to use.

Work done is equal to the dot or scalar product of the force and displacement vectors. The power 𝑃 is this expression differentiated with respect to time such that 𝑃 is equal to dπ‘Š by d𝑑. The power can also be calculated by finding the dot product of the force and velocity vectors. If we are given an expression for the work done, we can differentiate to find an expression for the power. And since integration is the opposite of differentiation, work done is the integral of power. Finally, we saw that the average power is equal to the work done divided by the change in time. We will now look at some examples of this.

The work done by an engine at time 𝑑 is given by the relation π‘Š of 𝑑 is equal to two 𝑑 cubed six 𝑑 joules. Find the power of the engine as a function of time.

We begin by recalling that power is the derivative of work such that 𝑃 is equal to dπ‘Š by d𝑑. In this question, we’re given an expression for the work done π‘Š in terms of time. It is equal to two 𝑑 cubed plus six 𝑑. We can therefore differentiate this function term by term to find a function for the power 𝑃. Recalling that if 𝑦 is equal to π‘Ž multiplied by π‘₯ to the power of 𝑛, then d𝑦 by dπ‘₯ is equal to 𝑛 multiplied by π‘Ž multiplied by π‘₯ to the power of 𝑛 minus one. We can differentiate two 𝑑 cubed to give us six 𝑑 squared as three multiplied by two is six, and we decrease the power by one. Differentiating six 𝑑 simply gives us six as when we decrease the power, we get 𝑑 to the power of zero, which is equal to one. The power of the engine as a function of time is equal to six 𝑑 squared plus six with the standard unit of watts.

In our next question, we need to find the power of a particle moving under a force in vector form.

A particle is moving under the action of the force 𝐅 is equal to eight 𝐒 plus five 𝐣. Its position vector at time 𝑑 is given by the relation 𝐫 of 𝑑 is equal to three 𝑑 squared minus 𝑑 minus four 𝐒 plus five over two 𝑑 squared plus 10𝑑 plus five 𝐣. Find the rate of work done by the force at 𝑑 equals three.

In this question, we’re trying to find the rate of the work done. This is the same as being asked to calculate the power since 𝑃 is equal to dπ‘Š by d𝑑. We can calculate the power by differentiating the work done with respect to time. In the question, we are given expressions for the vector force and displacement of the particle, noting that 𝐬 and 𝐫 are interchangeable for the displacement. Recalling that the work done is equal to the dot product of these, we will work this out first. We need to find the dot product of eight 𝐒 plus five 𝐣 and three 𝑑 squared minus 𝑑 minus four 𝐒 plus five over two 𝑑 squared plus 10𝑑 plus five 𝐣.

In order to calculate the dot product, we multiply the 𝐒- and 𝐣-components and then find the sum of these expressions. Eight multiplied by three 𝑑 squared minus 𝑑 minus four gives us 24𝑑 squared minus eight 𝑑 minus 32. Repeating this process for the 𝐣-components gives us 25 over two 𝑑 squared plus 50𝑑 plus 25. Collecting like terms, this expression simplifies to 73 over two 𝑑 squared plus 42𝑑 minus seven. We can now calculate the rate of the work done by differentiating this. Differentiating term by term gives us 73𝑑 plus 42, recalling that differentiating a constant gives us zero.

We are asked to find this rate of work done or power when 𝑑 is equal to three. Substituting this into our expression, we have 73 multiplied by three plus 42. This is equal to 261. The rate of the work done by the force at 𝑑 equals three is 261 units of power.

In our final example, we need to find the work done given an expression for the power in terms of 𝑑.

The power of an engine at time 𝑑 seconds is given by 𝑃 of 𝑑 is equal to 12𝑑 squared plus eight 𝑑 watts. Find the work done by the engine between 𝑑 equals two seconds and 𝑑 equals three seconds.

In this question, we are given an expression for the power of an engine in terms of 𝑑. We recall that power is the derivative of work such that 𝑃 is equal to dπ‘Š by d𝑑. As integration is the opposite or inverse of differentiation, the work must be the integral of the power with respect to 𝑑. We can therefore find an expression for the work done by integrating 12𝑑 squared plus eight 𝑑 with respect to 𝑑. As we need to find the work done between 𝑑 equals two and 𝑑 equals three seconds, our lower and upper limits will be two and three, respectively. This can also be demonstrated graphically as shown. The work done will be equal to the area bounded by the curve the vertical lines 𝑑 equals two and 𝑑 equals three and the horizontal axis.

We can integrate the expression for power term by term. We recall that the integral of π‘Žπ‘₯ to the power of 𝑛 with respect to π‘₯ is equal to π‘Žπ‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one where 𝑛 is not equal to negative one. In this question, since we’re dealing with a definite integral, we will not need to include the constant 𝐢. Integrating 12𝑑 squared gives us 12𝑑 cubed over three, which simplifies to four 𝑑 cubed. Integrating eight 𝑑 gives us eight 𝑑 squared over two, which simplifies to four 𝑑 squared.

Our next step is to substitute our upper and lower limits. This gives us four multiplied by three cubed plus four multiplied by three squared minus four multiplied by two cubed plus four multiplied by two squared. This in turn simplifies to 144 minus 48, which is equal to 96. Since the power of the engine was measured in the standard unit of watts, the work will be measured in joules. The work done by the engine between 𝑑 equals two and 𝑑 equals three seconds is 96 joules.

We will now summarize the key points from this video. We saw in this video that power is the rate at which a force does work on a body. This means that the power is the derivative of work such that 𝑃 is equal to dπ‘Š by d𝑑. We saw that the standard units of power are watts and the standard units of work are joules. Since integration is the inverse of differentiation, we saw that work is the integral of power. By using both of these formulae, we were able to calculate the amount of work done or power over a time period. We also saw that we can calculate the work done by finding the dot or scalar product of the force and displacement vectors and the power by finding the dot or scalar product of the force and velocity vectors.

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