Question Video: Using Vectors to Find the Work Done by a Force with the Direction and Magnitude Given Separately Mathematics

Video Transcript

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.