# Lesson Video: Work and Integration Mathematics

In this video, we will learn how to use integration to find the work done by a variable force.

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### 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.