### 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 is equal to the dot or scalar product of π
and π¬, where π
and π¬ 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 ππ£ 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 π is therefore equal to
dπ by dπ‘. We differentiate the work done with
respect to time. Letβs now consider how this works
in practice.

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

The rate at which work done
increases between π‘ equals zero and the instant π‘ is equal to the gradient of the
curve at the instant π‘. Recalling the power is equal to dπ
by dπ‘, and differentiating with respect to π‘, we have dπ by dπ‘ is equal to the
dot product of π
and ππ‘. If the force is not constant, it
does not produce a constant acceleration. So the instantaneous power π can
be given by the more general form the dot product of π
and π―, where vector π― 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 Ξπ‘. On our workβtime graph, the average
power is therefore equal to the gradient of the straight line that intersects the
value of π at the initial and final time values of the interval in the graph shown
from π‘ equals zero to π‘ equals five. Before looking at some specific
questions, letβs 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 π is this expression
differentiated with respect to time such that π is equal to dπ by dπ‘. 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
π‘ is given by the relation π of π‘ is equal to two π‘ cubed six π‘ 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 π is equal to dπ by dπ‘. In this question, weβre given an
expression for the work done π in terms of time. It is equal to two π‘ cubed plus
six π‘. We can therefore differentiate this
function term by term to find a function for the power π. Recalling that if π¦ is equal to π
multiplied by π₯ to the power of π, then dπ¦ by dπ₯ is equal to π multiplied by π
multiplied by π₯ to the power of π minus one. We can differentiate two π‘ cubed
to give us six π‘ squared as three multiplied by two is six, and we decrease the
power by one. Differentiating six π‘ simply gives
us six as when we decrease the power, we get π‘ to the power of zero, which is equal
to one. The power of the engine as a
function of time is equal to six π‘ 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 π
is equal to eight π’ plus five π£. Its position vector at time π‘ is
given by the relation π« of π‘ is equal to three π‘ squared minus π‘ minus four π’
plus five over two π‘ squared plus 10π‘ plus five π£. Find the rate of work done by the
force at π‘ equals three.

In this question, weβre trying to
find the rate of the work done. This is the same as being asked to
calculate the power since π is equal to dπ by dπ‘. 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 π¬
and π« 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 π’ plus five π£ and three π‘ squared minus π‘ minus four π’ plus five over two
π‘ squared plus 10π‘ plus five π£.

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

We are asked to find this rate of
work done or power when π‘ 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 π‘ 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 π‘.

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

In this question, we are given an
expression for the power of an engine in terms of π‘. We recall that power is the
derivative of work such that π is equal to dπ by dπ‘. As integration is the opposite or
inverse of differentiation, the work must be the integral of the power with respect
to π‘. We can therefore find an expression
for the work done by integrating 12π‘ squared plus eight π‘ with respect to π‘. As we need to find the work done
between π‘ equals two and π‘ 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 π‘ equals two and π‘ equals three and
the horizontal axis.

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