Lesson Video: Work Done by a Force Expressed in Vector Notation | Nagwa Lesson Video: Work Done by a Force Expressed in Vector Notation | Nagwa

Lesson Video: Work Done by a Force Expressed in Vector Notation Mathematics

In this video, we will learn how to calculate the work done by a constant force vector acting on a body over a displacement vector using the dot product.

13:35

Video Transcript

In this lesson, we’ll learn how to calculate the work done by a constant force vector acting on a body over a displacement vector using the dot product.

Remember, work is done when a force causes a body to move over some distance. And in its most simplest form, it can be calculated by finding the product of those two measures, force times displacement. When the force 𝐹 is constant and the angle between the force and the displacement 𝑑 is 𝜃, then work done is alternatively given by 𝐹 times 𝑑 cos 𝜃. And whilst work is a scalar quantity, we can alternatively represent the force and displacement as vectors rather than as the magnitude of vectors.

Then the work done by a force vector 𝐅 with displacement vector 𝐝 is the dot product of the force vector and the displacement vector. This can be alternatively calculated using the magnitude of the force vector times the magnitude of the displacement vector times cos 𝜃. And of course since the dot product or scalar product of two vectors is a scalar quantity, then this definition holds with our understanding of how work is represented. So, given these definitions, we’ll consider a demonstration of how to calculate work done.

A particle moves in a plane in which 𝐢 and 𝐣 are perpendicular unit vectors. A force 𝐅 equals nine 𝐢 plus 𝐣 newtons acts on the particle. The particle moves from the origin to the point with position vector negative nine 𝐢 plus six 𝐣 meters. Find the work done by the force.

Remember, the work done by a force vector 𝐅 with displacement vector 𝐝 is the dot product of 𝐅 and 𝐝. The force acting on our particle is nine 𝐢 plus 𝐣 newtons. But what is its displacement? Well, we’re told it moves from the origin to the point with position vector — let’s call that 𝐩 — negative nine 𝐢 plus six 𝐣. This means that the displacement is simply the vector 𝐨𝐩. That’s negative nine 𝐢 plus six 𝐣 minus zero 𝐢 plus zero 𝐣. And of course, we simply subtract the components. In this case, of course, we’re subtracting the vector zero 𝐢 plus zero 𝐣. So, our original vector remains unchanged. And so the displacement, which we can give in meters, is negative nine 𝐢 plus six 𝐣. So, work done is the dot product or the scalar product of 𝐅 and 𝐝. That’s nine 𝐢 plus 𝐣 dot negative nine 𝐢 plus six 𝐣.

Then, of course, we find the scalar product by finding the sum of the products of the components. So, we multiply the 𝐢-components, nine times negative nine, and then we add the product of the 𝐣-components. Well, that’s one times six. That’s negative 81 plus six, which is, of course, negative 75. Since the force is in newtons and the displacement is in meters, the work done is going to be in joules. Now, we notice that the work done is negative. Well, if the energy of the particle is conserved, then the kinetic energy of the particle will decrease. If the energy of the particle is not conserved, the work done might instead increase the potential energy of the particle. Either way, the work done by the force in this question is negative 75 joules.

In this question, we considered a single force acting upon a body. But there will be times where multiple forces are acting on the same body. In these cases, we’ll consider the resultant of these forces. And that will help us to calculate the work done. So, in our second example, we’re going to demonstrate what that looks like.

A body moves in a plane in which 𝐢 and 𝐣 are perpendicular unit vectors. Two forces 𝐅 sub one equals nine 𝐢 minus two 𝐣 newtons and 𝐅 sub two equals nine 𝐢 minus seven 𝐣 newtons act on the body. The particle moves from the point with position vector negative six 𝐢 plus two 𝐣 meters to the point two 𝐢 plus three 𝐣 meters. Find the work done by the resultant of the forces.

We begin by recalling that we can find the work done by a force vector 𝐅 over some displacement vector 𝐝 by finding the dot product of these two vectors. We do need to be a bit careful here though, because we’re working with two forces and we want to find the work done by their resultant. Now, of course, the resultant of two forces is simply their sum. So, the resultant force 𝐅 is the sum of 𝐅 sub one and 𝐅 sub two. In this case, that’s nine 𝐢 minus two 𝐣 plus nine 𝐢 minus seven 𝐣. Now, of course, we simply add the components, so we’ll add the 𝐢-components first. Nine plus nine is 18. Then, we add the 𝐣-components. Negative two minus seven is negative nine. And so, the resultant of the forces in this case is 18𝐢 minus nine 𝐣 newtons.

Now, we need to find the displacement. Now, of course, the particle moves from the point with position vector negative six 𝐢 plus two 𝐣 to the point two 𝐢 plus three 𝐣. Let’s call the first point 𝐴 and the second point 𝐵. Then, the displacement is the vector 𝐀𝐁. Well, that’s the difference between vector 𝐎𝐁 and 𝐎𝐀. So, it’s two 𝐢 plus three 𝐣 minus negative six 𝐢 plus two 𝐣. As before, we simply subtract the individual components, beginning with the 𝐢-components. Two minus negative six is eight, and three minus two is one. So, we find the displacement of the body is eight 𝐢 plus 𝐣. And of course, that’s in meters.

And we have everything we need. The work done is simply the dot product or the scalar product of these two vectors. That’s 18 times eight plus negative nine times one, which is 144 minus nine, or 135. The work done by the resultant of the two forces then is 135 joules.

In our next example, we’ll have a look at how to use our understanding of the features of vectors to find work done by a force.

A particle moved from point 𝐴 seven, negative three to point 𝐵 negative nine, two along a straight line under the action of a force 𝐅 of magnitude eight root 10 newtons acting in the same direction as the vector 𝐜 equals negative three 𝐢 minus 𝐣. Calculate the work done by the force, given that the magnitude of the displacement is measured in meters.

Remember, the work done by a force of vector 𝐅 over some displacement vector 𝐝 is the dot product or scalar product of those two vectors. In this question, we’re told that the particle moves from point 𝐴 to point 𝐵, so we can calculate the displacement of the particle fairly easily. In vector form, we’ll say that the displacement 𝐝 is the vector 𝐀𝐁, and that’s the vector 𝐎𝐁 minus the vector 𝐎𝐀. Now, since we know the position from the origin, that’s the point zero, zero, then the vector 𝐎𝐁 is simply negative nine 𝐢 plus two 𝐣 and the vector 𝐎𝐀 is seven 𝐢 minus three 𝐣. So, the vector 𝐝 is the difference of these, and we can find that by subtracting the individual components. Negative nine minus seven is negative 16, and two minus negative three is five 𝐣. So, we have the displacement vector. It’s negative 16𝐢 plus five 𝐣.

How do we calculate the force though? We know it has a magnitude of eight root 10, but we don’t know its vector. We do, however, know it acts in the same direction as this vector here, negative three 𝐢 minus 𝐣. So, let’s sketch this out. Vector 𝐜 looks a little something like this. For every three units left, we move one unit down. Vector 𝐅 acts in the same direction as this but has a magnitude of eight root 10. Once again, for every three units left, we still move one unit down. So, we can say that if we define 𝐅 sub 𝑖 as being the 𝐢-component of our force and 𝐅 sub 𝑗 as being the 𝐣-component, then 𝐅 sub 𝑖 is equal to three times 𝐅 sub 𝑗.

And, of course, since the 𝐢-components and 𝐣-components of the vectors are perpendicular to one another, we can form a right triangle and apply the Pythagorean theorem to find the value of 𝐅 sub 𝑖 and 𝐅 sub 𝑗. That is, the magnitude of 𝐅 sub 𝑖 squared plus the magnitude of 𝐅 sub 𝑗 squared is equal to eight root 10 all squared. Now, of course, 𝐅 sub 𝑖 is three times 𝐅 sub 𝑗. So, we can replace the magnitude of 𝐅 sub 𝑖 with three times the magnitude of 𝐅 sub 𝑗. Similarly, eight root 10 all squared is 640. Three squared is nine, so we have a total of 10 lots of 𝐅 sub 𝑗 squared. Then, we divide through by 10. And finally, we’ll take the square root of both sides. Now, we only need to take the positive square root of 64 because we’re looking at the magnitude, which is simply a length. And so, the magnitude of 𝐅 sub 𝑗 is equal to eight. Since the magnitude of 𝐅 sub 𝑖 is three times the magnitude of 𝐅 sub 𝑗, it’s three times eight, which is equal to 24. We know the direction in which our force is acting. So we can define the vector force to be negative 24𝐢 minus eight 𝐣 newtons.

Let’s clear some space and we’re ready to take the dot product of our force and our displacement. The work done is the dot or scalar product of negative 24𝐢 minus eight 𝐣 and negative 16𝐢 plus five 𝐣. That’s negative 24 times negative 16 plus negative eight times five, and that’s equal to 344. The work done is in joules. So, we find the work done by our force is 344 joules.

In our previous examples, we’ve looked at finding the work done when both displacement and force are vectors, where each component is simply a number. In our final example, we’ll demonstrate how work done can also be calculated when the vectors for displacement and/or force are given as expressions in terms of time.

A particle moves in a plane in which 𝐢 and 𝐣 are perpendicular unit vectors. Its displacement from the origin at time 𝑡 seconds is given by 𝐫 equals two 𝑡 squared plus seven 𝐢 plus 𝑡 plus seven 𝐣 meters. And it is acted on by a force 𝐅 equals six 𝐢 plus three 𝐣 newtons. How much work does the force do between 𝑡 equals two seconds and 𝑡 equals three seconds?

Since work done is the dot product of the vector force and the vector for displacement, we’re going to need to calculate the displacement of this particle. We’re given its displacement at some time 𝑡. But we’re also asked how much work the force does between 𝑡 equals two seconds and 𝑡 equals three seconds. So, let’s substitute 𝑡 equals two and 𝑡 equals three into our expressions for the displacement. Beginning with 𝑡 equals three seconds, we get two times three squared plus seven 𝐢 plus three plus seven 𝐣. That’s 25𝐢 plus 10𝐣. And we’re working in meters. So the displacement from the origin and, hence, the position of the particle at three seconds is 25𝐢 plus 10𝐣 meters. In a similar way, we can find the position of the particle at two seconds by substituting 𝑡 equals two into our expression. And we get 15𝐢 plus nine 𝐣 meters.

The displacement then of the particle between these two times is the difference here. It’s 𝐝 sub three minus 𝐝 sub two. That’s 25𝐢 plus 10𝐣 minus 15𝐢 plus nine 𝐣. And of course, we simply subtract the individual components. 25 minus 15 is 10, and 10 minus nine is one. So, the displacement is simply 10𝐢 plus 𝐣, and that’s in meters. Since the work done is the dot product of the force and displacement, we need to find the dot product of six 𝐢 plus three 𝐣 and the displacement we just calculated. It’s the dot or scalar product of six 𝐢 plus three 𝐣 and 10𝐢 plus 𝐣. That’s six times 10 plus three times one, which is, of course, equal to 63. The work done then is equal to 63 joules.

We’ll now recap the key points from this lesson. In this lesson, we learned the work done by a constant force vector 𝐅 over a displacement vector 𝐝 is equal to the dot product of 𝐅 and 𝐝. Alternatively, if 𝜃 is the angle between 𝐅 and 𝐝, it’s the magnitude of the force times the magnitude of the displacement times cos 𝜃. And if we express 𝐅 and 𝐝 in component form, we don’t actually need to use this second form. We can simplify the dot product to find the work done.

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