Video Transcript
In this video, we will learn how to
use integration to find the work done by a variable force. We recall that for a constant force
𝐅 that acts on an object as that object undergoes a displacement 𝑠, the work done
by the force 𝑊 is the scalar product of the force and the displacement. This can be written as 𝑊 is equal
to the dot or scalar product of vector 𝐅 and vector 𝐬. This can also be written as 𝑊 is
equal to 𝐅 multiplied by 𝑠 multiplied by the cos of 𝜃, where 𝐅 is the magnitude
of the force, 𝑠 is the magnitude of the displacement, and 𝜃 is the angle between
the force acting on the object and its displacement.
Let’s now consider three different
scenarios of this. Firstly, we will consider what
happens when 𝐅 and 𝜃 are constant. If the magnitude of the force is
constant and the angle between the force and the displacement does not vary, then
the graph of 𝐅 cos 𝜃 against 𝑠 would look as shown. The value of 𝐅 cos 𝜃 remains
constant over the path the object takes. And the work done by the force is
equal to the area under the line. As this is a rectangular area, the
work done by the force is equal to the length multiplied by the width. This is equal to 𝑠 multiplied by
𝐅 cos 𝜃. This can also be calculated for
different regions of the graph, for example, between the points a and b on the
horizontal axis.
Now let’s consider what happens if
𝐅 varies as the object moves. For example, if 𝐅 increases before
reaching a constant value, then the graph of 𝐅 cos 𝜃 against 𝑠 might look
something like this. Now, in order to find the area
under the line, i.e., the work done, we have to divide the area into two
regions. We have a trapezoid or trapezium
and a rectangle. The work done would be equal to the
sum of the areas of the trapezoid and rectangle.
Finally, let’s consider what
happens if the force 𝐅 is described as a continuous function. We would now need to use
integration to find the area under the curve and hence the work done. Assuming that the force is a
function in terms of 𝑠, then the work done is equal to the integral of 𝐅 cos 𝜃
with respect to 𝑠. If the force and displacement are
in the same direction, then 𝜃 is equal to zero. And we know that the cos of zero is
equal to one. This means that our formula can be
simplified to 𝑊 is equal to the integral of 𝐅 with respect to 𝑠. This enables us to work out the
work done by a force on an object as the object moves on a path parallel to the
force. We will now look at some examples
where we need to calculate the work done by integration between two intervals a and
b.
A body moves along the 𝑥-axis
under the action of a force 𝐅. Given that 𝐅 is equal to eight 𝑠
plus 12 newtons, where 𝑠 meters is the displacement from the origin, determine the
work done on the body by 𝐅 when the body moves from 𝑠 equals seven meters to 𝑠
equals eight meters.
In this question, a variable force
acts on an object. And both the motion of the object
and the force acting on it are along the 𝑥-axis. We can therefore calculate the work
done by using the formula 𝑊 is equal to the integral of 𝐅 with respect to 𝑠. We are told that the force 𝐅 is
equal to eight 𝑠 plus 12. This means in order to calculate
the work done, we need to integrate this expression with respect to 𝑠. As we need to calculate the work
done between 𝑠 equals seven meters and 𝑠 equals eight meters, our lower limit is
seven and our upper limit is eight. Integrating eight 𝑠 gives us eight
𝑠 squared over two, which simplifies to four 𝑠 squared. Integrating 12 with respect to 𝑠
gives us 12𝑠. The work done is therefore equal to
four 𝑠 squared plus 12𝑠.
Our next step is to substitute the
limits. When 𝑠 is equal to eight, 𝑊 is
equal to 352. And when 𝑠 is equal to seven, 𝑊
is equal to 280. The work done over this distance is
therefore equal to 352 minus 280. This is equal to 72. Since the force 𝐅 was measured in
newtons and the displacement 𝑠 in meters, our work done will be in
newton-meters. We know that this is equivalent to
joules. Therefore, the work done on the
body by 𝐅 when the body moves from 𝑠 equals seven meters to 𝑠 equals eight meters
is 72 joules.
In our next example, the expression
for the force will involve a trigonometric function.
A particle moves in a straight line
under the action of the force 𝐅, where 𝐅 is equal to sin of 𝜋𝑠 and 𝑠 is
measured in meters. Calculate the work done by the
force 𝐅 when the particle moves from 𝑠 equals zero to 𝑠 equals one-half.
In this question, a variable force
acts on a particle. And both the motion of the particle
and the force acting on it are in one dimension. We can therefore calculate the work
done by the force by using the formula 𝑊 is equal to the integral of 𝐅 with
respect to 𝑠. We are told in the question that
the force 𝐅 is equal to sin of 𝜋𝑠. The work done is therefore equal to
the integral of this with respect to 𝑠. And we need to calculate this
between 𝑠 equals zero and 𝑠 equals one-half. So these are our lower and upper
limits.
We recall that the integral of sin
𝑎𝑥 with respect to 𝑥 is equal to negative one over 𝑎 multiplied by the cos of
𝑎𝑥. This means that our expression
integrates to negative one over 𝜋 multiplied by the cos of 𝜋𝑠. Our next step is to substitute in
our limits. When 𝑠 is equal to one-half, we
have negative one over 𝜋 multiplied by cos of 𝜋 over two. The cos of 𝜋 over two radians or
90 degrees is zero. This means that when 𝑠 equals
one-half, the work done equals zero. When 𝑠 is equal to zero, we have
negative one over 𝜋 multiplied by the cos of zero. As the cos of zero is one, we are
left with negative one over 𝜋.
The work done between our limits is
therefore equal to zero minus negative one over 𝜋. This simplifies to one over 𝜋. When our force is measured in
newtons and the displacement in meters, then the work done is measured in
newton-meters or joules. We can therefore conclude that the
work done by the force 𝐅 is one over 𝜋 joules.
In our final example, we will
calculate the work done by a variable force with an unknown constant.
A block moves in a straight line
under the action of a force 𝐅 equal to 12𝑠 squared plus six 𝑠 plus 𝑐 newtons,
where 𝑠 meters is the displacement of the body from its initial position. The work done by the force in
moving the block from 𝑠 equals zero meters to 𝑠 equals three meters is 34
joules. Determine the work done by 𝐅 in
moving the block from 𝑠 equals three meters to 𝑠 equals six meters.
As we have a force acting on an
object moving in a straight line, we can use the formula 𝑊 is equal to the integral
of 𝐅 with respect to 𝑠 to calculate the work done by the force. In this question, we are told that
the force 𝐅 is equal to 12𝑠 squared plus six 𝑠 plus 𝑐 newtons. The work done is therefore equal to
the integral of this with respect to 𝑠. Integrating each term with respect
to 𝑠, we have four 𝑠 cubed plus three 𝑠 squared plus 𝑐𝑠. We are told that the work done by
the force in moving the block from 𝑠 equals zero meters to 𝑠 equals three meters
is 34 joules. We can therefore substitute in
these values as shown as this will enable us to calculate the constant 𝑐.
When 𝑠 equals three, the
right-hand side of our equation becomes 135 plus three 𝑐. And when 𝑠 equals zero, this
expression equals zero. This means that 34 is equal to 135
plus three 𝑐. Subtracting 135 from both sides of
this equation, we have three 𝑐 is equal to negative 101. We can then divide through by three
such that 𝑐 is equal to negative 101 over three. Substituting this back in to the
expression for the work done, we have 𝑊 is equal to four 𝑠 cubed plus three 𝑠
squared minus 101 over three 𝑠.
As we need to calculate the work
done by 𝐅 from 𝑠 equals three to 𝑠 equals six, we can substitute these values
into our expression. When 𝑠 equals six 𝑊 is equal to
770, and when 𝑠 equals three, 𝑊 is equal to 34. The work done from 𝑠 equals three
meters to 𝑠 equals six meters is therefore equal to 770 minus 34, which is equal to
736. The final answer is equal to 736
joules.
We will now summarize the key
points from this video. We saw in this video that we can
use integration to find the work done on an object by a variable force. The work done by a force on an
object as the object moves along a path parallel to the force is given by 𝑊 is
equal to the integral of 𝐅 d𝑠, where 𝑊 is the work done, 𝐅 is the magnitude of
the force that acts on the object, and d𝑠 is an infinitesimal line segment of the
path.
