Lesson Explainer: Work and Integration | Nagwa Lesson Explainer: Work and Integration | Nagwa

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, 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: 𝑊=𝐹𝑠.

This can also be written as 𝑊=𝐹𝑠𝜃,cos 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.

If 𝐹 and 𝜃 are constant—in other words, the magnitude of the force is constant and the angle between the force and the displacement does not vary—the graph of 𝐹𝜃cos against 𝑠 would look like this:

𝐹𝜃cos remains constant over the path that the object takes. The work done by the force, 𝑊, 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: 𝐹𝜃×𝑠cos, or 𝐹𝑠𝜃cos.

Now, let’s imagine that 𝐹 varies as the object moves. Let’s imagine that, at first, 𝐹 increases before reaching a constant value. The graph of 𝐹𝜃cos against 𝑠 might then look like this:

Now in order to find the area under the line—the work done—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 area—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 𝐹𝜃cos with respect to 𝑠: 𝑊=(𝐹𝜃)𝑠.cosd

If the force and displacement are in the same direction, 𝜃 is 0 and cos𝜃 is 1, which means that this formula can be simplified to 𝑊=𝐹𝑠.d

While our original definition for the work done by a force, 𝑊=𝐹𝑠𝜃cos, 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 𝑊=𝐹𝑠,d 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.

Let’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 𝑥-axis under the action of a force, 𝐹. Given that 𝐹=(8𝑠+12)N, where 𝑠 m is the displacement from the origin, determine the work done on the body by 𝐹 when the body moves from 𝑠=7m to 𝑠=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 𝑥-axis, so we can use the formula 𝑊=𝐹𝑠d to find the work done on the object.

We are asked to find the work done on the object when it moves between 𝑠=7m and 𝑠=8m, which means that this is going to be a definite integral with these values as the limits. Let’s start by substituting 𝐹, as well as these limits: 𝑊=(8𝑠+12)𝑠.d

By integrating 8𝑠+12, we get 𝑊=4𝑠+12𝑠, and then if we resolve this expression between the two limits, we get 𝑊=4×(8)+12×(8)4×(7)+12×(7)𝑊=(352)(280)𝑊=72.

Since 𝐹 was measured in newtons and 𝑠 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 𝐹, measured in newtons, is acting on a body, where 𝐹=3𝑠5. Find the work done by this force in the interval from 𝑠=4m to 𝑠=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 𝑊=𝐹𝑠d to find the work done on the object.

We are asked to find the work done on the object when it moves between 𝑠=4m and 𝑠=5m. To do this, we can use a definite integral with these values as the limits. Let’s start by substituting 𝐹, as well as these limits: 𝑊=3𝑠5𝑠.d

By integrating 3𝑠5, we get 𝑊=𝑠5𝑠, and then if we resolve this expression between the two limits, we get 𝑊=55×545×4𝑊=(100)(44)𝑊=56.

Since 𝐹 was measured in newtons and 𝑠 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 𝐹, where 𝐹=𝜋𝑆sin and 𝑆 is measured in metres. Calculate the work done by the force 𝐹 when the particle moves from 𝑆=0 to 𝑆=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 𝑊=𝐹𝑆d to find the work done on the particle.

We are asked to find the work done on the particle when it moves between 𝑆=0m and 𝑆=12m. To do this, we can use a definite integral with these values as the limits. Let’s start by substituting 𝐹, as well as these limits: 𝑊=(𝜋𝑆)𝑆.sind

By integrating sin𝜋𝑆, we get 𝑊=1𝜋𝜋𝑆,cos and then if we resolve this expression between the two limits, we get 𝑊=1𝜋𝜋×121𝜋(𝜋×0)𝑊=1𝜋𝜋21𝜋(0).coscoscoscos

The cosine of zero is one, and the cosine of 𝜋2 is zero, so 𝑊=1𝜋×01𝜋𝑊=1𝜋𝑊=1𝜋.

Since 𝐹 was measured in newtons and 𝑆 was measured in metres, this value for the work is in newton-metres, which is equivalent to joules, so the work done is 1𝜋 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 𝐹=12𝑠+6𝑠+𝑐N, where 𝑠 metres is the displacement of the body from its initial position. The work done by the force in moving the block from 𝑠=0m to 𝑠=3m is 34 J. Determine the work done by 𝐹 in moving the block from 𝑠=3m to 𝑠=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 𝑊=𝐹𝑠d to find the work done on the object.

We are asked to find the work done on the object when it moves between 𝑠=3m and 𝑠=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, 𝑐, 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, 𝑐. Let’s start by working out this integral: 34=12𝑠+6𝑠+𝑐𝑠.d

By integrating 12𝑠+6𝑠+𝑐, we get 34=4𝑠+3𝑠+𝑐𝑠, and then if we resolve this expression between the two limits, we get 34=4×(3)+3×(3)+3𝑐4×(0)+3×(0)+0𝑐34=(4×27+3×9+3𝑐)34=135+3𝑐.

We can then rearrange the equation to make 𝑐 the subject: 34135=3𝑐101=3𝑐3𝑐=101𝑐=1013.

This means that we now have a complete expression for 𝐹: 𝐹=12𝑠+6𝑠1013.N

From here, we can write out a definite integral for the work done between 𝑠=3m and 𝑠=6m: 𝑊=12𝑠+6𝑠1013𝑠,d and then we just follow the same steps as we did before: 𝑊=4𝑠+3𝑠1013𝑠𝑊=4×(6)+3×(6)1013×64×(3)+3×(3)1013×3𝑊=77034𝑊=736.J

The work done by the force between 𝑠=3m and 𝑠=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 𝑊=𝐹𝑠,d 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.

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