Lesson Explainer: Work and Integration Mathematics

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

Recall that for a constant force, โƒ‘๐น, that acts on an object as that object undergoes a displacement, โƒ‘๐‘ , the work done by the force, ๐‘Š, is the scalar product of the force and the displacement: ๐‘Š=โƒ‘๐นโ‹…โƒ‘๐‘ .

This can also be written as ๐‘Š=๐น๐‘ ๐œƒ,cos where ๐น is the magnitude of the force, ๐‘  is the magnitude of the displacement, and ๐œƒ is the angle between the force acting on the object and its displacement.

If ๐น and ๐œƒ are constantโ€”in other words, the magnitude of the force is constant and the angle between the force and the displacement does not varyโ€”the graph of ๐น๐œƒcos against ๐‘  would look like this:

๐น๐œƒcos remains constant over the path that the object takes. The work done by the force, ๐‘Š, is equal to the area under the line. The area under the line is just a single rectangular region, so the area is equal to the height of the rectangle multiplied by its width: ๐น๐œƒร—๐‘ cos, or ๐น๐‘ ๐œƒcos.

Now, letโ€™s imagine that ๐น varies as the object moves. Letโ€™s imagine that, at first, ๐น increases before reaching a constant value. The graph of ๐น๐œƒcos against ๐‘  might then look like this:

Now in order to find the area under the lineโ€”the work doneโ€”we will have to divide the area into two regions, a trapezium and a rectangle, and find the area of each of those.

We can see that as the force on the object becomes more complex, we would need to divide the area under the line into more regions in order to be able to calculate the total areaโ€”the work done. If the force acting on an object were described by a continuous function, such as in the graph below, we would have to use integration to find the area under the curve and, hence, the work done.

In order to find the area under the curve in the graph above, we would have to take the integral of ๐น๐œƒcos with respect to ๐‘ : ๐‘Š=๏„ธ(๐น๐œƒ)๐‘ .cosd

If the force and displacement are in the same direction, ๐œƒ is 0 and cos๐œƒ is 1, which means that this formula can be simplified to ๐‘Š=๏„ธ๐น๐‘ .d

While our original definition for the work done by a force, ๐‘Š=๐น๐‘ ๐œƒcos, was valid for constant forces, we have come up with a formula for the work done by a variable force, as long as the force is in the same direction as the displacement. We have come up with a more general definition for the work done by a force.

Definition:

The work done by a force on an object as the object moves along a path parallel to the force is given by ๐‘Š=๏„ธ๐น๐‘ ,d where ๐‘Š is the work done, ๐น is the magnitude of the force that acts on the object, and d๐‘  is an infinitesimal line segment of the path.

Letโ€™s have a look at some examples.

Example 1: Calculating the Amount of Work Done by a Force given the Force Expression

A body moves along the ๐‘ฅ-axis under the action of a force, ๐น. Given that ๐น=(8๐‘ +12)N, where ๐‘  m is the displacement from the origin, determine the work done on the body by ๐น when the body moves from ๐‘ =7m to ๐‘ =8m.

Answer

In this question, a variable force acts on an object. Both the motion of the object and the force acting on it are along the ๐‘ฅ-axis, so we can use the formula ๐‘Š=๏„ธ๐น๐‘ d to find the work done on the object.

We are asked to find the work done on the object when it moves between ๐‘ =7m and ๐‘ =8m, which means that this is going to be a definite integral with these values as the limits. Letโ€™s start by substituting ๐น, as well as these limits: ๐‘Š=๏„ธ(8๐‘ +12)๐‘ .๏Šฎ๏Šญd

By integrating 8๐‘ +12, we get ๐‘Š=๏‘4๐‘ +12๐‘ ๏,๏Šจ๏Šฎ๏Šญ and then if we resolve this expression between the two limits, we get ๐‘Š=๏€น4ร—(8)+12ร—(8)๏…โˆ’๏€น4ร—(7)+12ร—(7)๏…๐‘Š=(352)โˆ’(280)๐‘Š=72.๏Šจ๏Šจ

Since ๐น was measured in newtons and ๐‘  was measured in metres, this value for the work is in newton-metres, which is equivalent to joules, so the work done is 72 J.

Example 2: Calculating the Amount of Work Done by a Force given the Force Expression

A variable force ๐น, measured in newtons, is acting on a body, where ๐น=3๐‘ โˆ’5๏Šจ. Find the work done by this force in the interval from ๐‘ =4m to ๐‘ =5m.

Answer

In this question, a variable force acts on an object. Both the motion of the object and the force acting on it are in one dimension, so we can use the formula ๐‘Š=๏„ธ๐น๐‘ d to find the work done on the object.

We are asked to find the work done on the object when it moves between ๐‘ =4m and ๐‘ =5m. To do this, we can use a definite integral with these values as the limits. Letโ€™s start by substituting ๐น, as well as these limits: ๐‘Š=๏„ธ๏€น3๐‘ โˆ’5๏…๐‘ .๏Šซ๏Šช๏Šจd

By integrating 3๐‘ โˆ’5๏Šจ, we get ๐‘Š=๏‘๐‘ โˆ’5๐‘ ๏,๏Šฉ๏Šซ๏Šช and then if we resolve this expression between the two limits, we get ๐‘Š=๏€น5โˆ’5ร—5๏…โˆ’๏€น4โˆ’5ร—4๏…๐‘Š=(100)โˆ’(44)๐‘Š=56.๏Šฉ๏Šฉ

Since ๐น was measured in newtons and ๐‘  was measured in metres, this value for the work is in newton-metres, which is equivalent to joules, so the work done is 56 J.

Example 3: Calculating the Amount of Work Done by a Force That varies Sinusoidally

A particle moves in a straight line under the action of the force ๐น, where ๐น=๐œ‹๐‘†sin and ๐‘† is measured in metres. Calculate the work done by the force ๐น when the particle moves from ๐‘†=0 to ๐‘†=12.

Answer

In this question, a variable force acts on a particle. Both the motion of the particle and the force acting on it are in one dimension, so we can use the formula ๐‘Š=๏„ธ๐น๐‘†d to find the work done on the particle.

We are asked to find the work done on the particle when it moves between ๐‘†=0m and ๐‘†=12m. To do this, we can use a definite integral with these values as the limits. Letโ€™s start by substituting ๐น, as well as these limits: ๐‘Š=๏„ธ(๐œ‹๐‘†)๐‘†.๏Ž ๏Žก๏Šฆsind

By integrating sin๐œ‹๐‘†, we get ๐‘Š=๏”โˆ’1๐œ‹๐œ‹๐‘†๏ ,cos๏Ž ๏Žก๏Šฆ and then if we resolve this expression between the two limits, we get ๐‘Š=๏€ผโˆ’1๐œ‹๏€ผ๐œ‹ร—12๏ˆ๏ˆโˆ’๏€ผโˆ’1๐œ‹(๐œ‹ร—0)๏ˆ๐‘Š=๏€ผโˆ’1๐œ‹๏€ป๐œ‹2๏‡๏ˆโˆ’๏€ผโˆ’1๐œ‹(0)๏ˆ.coscoscoscos

The cosine of zero is one, and the cosine of ๐œ‹2 is zero, so ๐‘Š=๏€ผโˆ’1๐œ‹ร—0๏ˆโˆ’๏€ผโˆ’1๐œ‹๏ˆ๐‘Š=โˆ’๏€ผโˆ’1๐œ‹๏ˆ๐‘Š=1๐œ‹.

Since ๐น was measured in newtons and ๐‘† was measured in metres, this value for the work is in newton-metres, which is equivalent to joules, so the work done is 1๐œ‹ J.

Example 4: Calculating the Amount of Work Done by a Variable Force with an Unknown Constant

A block moves in a straight line under the action of a force ๐น=๏€น12๐‘ +6๐‘ +๐‘๏…๏ŠจN, where ๐‘  metres is the displacement of the body from its initial position. The work done by the force in moving the block from ๐‘ =0m to ๐‘ =3m is 34 J. Determine the work done by ๐น in moving the block from ๐‘ =3m to ๐‘ =6m.

Answer

In this question, a variable force acts on an object. Both the motion of the object and the force acting on it are in one dimension, so we can use the formula ๐‘Š=๏„ธ๐น๐‘ d to find the work done on the object.

We are asked to find the work done on the object when it moves between ๐‘ =3m and ๐‘ =6m, which means that we are going to need to work out a definite integral of the force on the object with these values as the limits. However, there is an unknown constant in the equation for the force, ๐‘, that we need to work out first; otherwise, this unknown would appear in the result of our definite integral and we would not be able to evaluate the work done.

In order to find the value of this constant, we can find the definite integral of the force between the limits of 0 m and 3 m. Since the value for the work between these limits has been given to us, we will be able to form an equation with only one unknown, ๐‘. Letโ€™s start by working out this integral: 34=๏„ธ๏€น12๐‘ +6๐‘ +๐‘๏…๐‘ .๏Šฉ๏Šฆ๏Šจd

By integrating 12๐‘ +6๐‘ +๐‘๏Šจ, we get 34=๏‘4๐‘ +3๐‘ +๐‘๐‘ ๏,๏Šฉ๏Šจ๏Šฉ๏Šฆ and then if we resolve this expression between the two limits, we get 34=๏€น4ร—(3)+3ร—(3)+3๐‘๏…โˆ’๏€น4ร—(0)+3ร—(0)+0๐‘๏…34=(4ร—27+3ร—9+3๐‘)34=135+3๐‘.๏Šฉ๏Šจ๏Šฉ๏Šจ

We can then rearrange the equation to make ๐‘ the subject: 34โˆ’135=3๐‘โˆ’101=3๐‘3๐‘=โˆ’101๐‘=โˆ’1013.

This means that we now have a complete expression for ๐น: ๐น=๏€ผ12๐‘ +6๐‘ โˆ’1013๏ˆ.๏ŠจN

From here, we can write out a definite integral for the work done between ๐‘ =3m and ๐‘ =6m: ๐‘Š=๏„ธ๏€ผ12๐‘ +6๐‘ โˆ’1013๏ˆ๐‘ ,๏Šฌ๏Šฉ๏Šจd and then we just follow the same steps as we did before: ๐‘Š=๏”4๐‘ +3๐‘ โˆ’1013๐‘ ๏ ๐‘Š=๏€ผ4ร—(6)+3ร—(6)โˆ’1013ร—6๏ˆโˆ’๏€ผ4ร—(3)+3ร—(3)โˆ’1013ร—3๏ˆ๐‘Š=770โˆ’34๐‘Š=736.๏Šฉ๏Šจ๏Šฌ๏Šฉ๏Šฉ๏Šจ๏Šฉ๏ŠจJ

The work done by the force between ๐‘ =3m and ๐‘ =6m is 736 J.

Key Points

  • 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 ๐‘Š=๏„ธ๐น๐‘ ,d where ๐‘Š is the work done, ๐น is the magnitude of the force that acts on the object, and d๐‘  is an infinitesimal line segment of the path.

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