### Video Transcript

Consider a particle on which several forces act, one of which is known to be constant in time: π
sub one is equal to 3.0π’ plus 4.0π£ newtons. As a result, the particle moves along a straight path from a Cartesian coordinate of zero metres, zero metres to 5.0 metres, 6.0 metres. What is the work done by π
sub one?

Letβs start by highlighting some of the critical information weβve been given. First, weβre told about the force called πΉ sub one. And weβre given the expression of that force in its π₯ and π¦ or π and π components. Then, weβre told about a particle that moves under the influence of at least πΉ sub one from the location starting at the origin zero metres, zero metres moving to the point 5.0 metres, 6.0 metres in a horizontal and a vertical plane.

Weβre looking for the work done by πΉ one on this particle over the journey, and weβll represent that work using the symbol capital π. As we begin, letβs recall a definition that relates work, force, and displacement. Work, capital π, is defined as a force exerted on an object multiplied by the displacement of that object. As we look over the problem statement, we see weβve been given a force; thatβs, πΉ sub one, and weβve also been given a displacement β the path that the particle travels. We can use this information then to solve for π, the work done by πΉ one.

Now as we look at our problem, we recognize it as a two-dimensional problem. Thereβs both a horizontal and a vertical component to πΉ one and also a horizontal and a vertical component to the motion of the particle. That means we can write our total work π as the sum of the work in the horizontal and the vertical directions. Weβll call those values of work π sub π₯ for the horizontal work done and π sub π¦ for the vertical work done.

If we refer to our definition for work, we can write out equations for work in the π₯- and π¦-directions: π sub π₯ is equal to πΉ sub π₯ times π sub π₯. This is a way of saying that the work in the horizontal direction is equal to the force in the horizontal direction times the displacement in the horizontal direction. Likewise, we can write a similar equation for π sub π¦, the work done in the vertical direction.

Now that we have these two equations for work in the horizontal and vertical directions, we can begin to fill in πΉ sub π₯ and π sub π₯ and πΉ sub π¦ and π sub π¦ from the given information in our problem. Take another look at πΉ sub one, the component in the π or horizontal direction is 3.0 newtons. So we can write that in for πΉ sub π₯.

Then, as we look at the particle path, and specifically at π point, we see that its displacement in the horizontal or π₯-direction is 5.0 metres minus 0 metres, which equals 5.0 metres. Now moving on to work in the π¦ or vertical direction, we see that the component of πΉ sub one that is vertical is 4.0 newtons. And the displacement of our particle in the π¦-direction or vertical direction is 6.0 metres minus zero metres or 6.0 metres.

We can now multiply these pairs of numbers together: 3.0 newtons times 5.0 metres equals 15 newton metres. And then for π sub π¦, 4.0 newtons times 6.0 metres equals 24 newton metres.

Now, we can use these horizontal and vertical components of the total work done to solve for π, the net work done on the particle by πΉ one. To do this, we need to sum π sub π₯ and π sub π¦: 15 newton metres plus 24 newton metres equals 39 newton metres. Another name for a newton metre is a joule. This is the total amount of work done on the particle by πΉ sub one as the particle moves along this straight line path.