# Video: Calculating the Work Done by a Position Dependent Force on a Moving Particle

Consider a particle on which several forces act, one of which is known to be constant in time: πβ = (3.0π’ + 4.0π£) N. As a result, the particle moves along a straight path from a Cartesian coordinate of (0 m, 0 m) to (5.0 m, 6.0 m). What is the work done by πβ?

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### 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?

Letβs start by highlighting some of the critical information weβve been given. First, weβre told about the force called πΉ sub one. And weβre given the expression of that force in its π₯ and π¦ or π and π components. Then, weβre told about a particle that moves under the influence of at least πΉ 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.

Weβre looking for the work done by πΉ one on this particle over the journey, and weβll represent that work using the symbol capital π. As we begin, letβs 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 weβve been given a force; thatβs, πΉ sub one, and weβve 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 πΉ one.

Now as we look at our problem, we recognize it as a two-dimensional problem. Thereβs both a horizontal and a vertical component to πΉ 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. Weβll 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 πΉ 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 πΉ sub π₯ and π sub π₯ and πΉ sub π¦ and π sub π¦ from the given information in our problem. Take another look at πΉ sub one, the component in the π or horizontal direction is 3.0 newtons. So we can write that in for πΉ 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 πΉ 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 πΉ 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 πΉ sub one as the particle moves along this straight line path.