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: 𝐅₁ = (3.0𝐒 + 4.0𝐣) 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 𝐅₁?

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Video Transcript

Consider a particle on which several forces act, one of which is known to be constant in time: 𝐅 sub one is equal to 3.0𝐒 plus 4.0𝐣 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 𝐅 sub one?

Let’s start by highlighting some of the critical information we’ve been given. First, we’re told about the force called 𝐹 sub one. And we’re given the expression of that force in its π‘₯ and 𝑦 or 𝑖 and 𝑗 components. Then, we’re told about a particle that moves under the influence of at least 𝐹 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.

We’re looking for the work done by 𝐹 one on this particle over the journey, and we’ll represent that work using the symbol capital π‘Š. As we begin, let’s recall a definition that relates work, force, and displacement. Work, capital π‘Š, 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 we’ve been given a force; that’s, 𝐹 sub one, and we’ve also been given a displacement β€” the path that the particle travels. We can use this information then to solve for π‘Š, the work done by 𝐹 one.

Now as we look at our problem, we recognize it as a two-dimensional problem. There’s both a horizontal and a vertical component to 𝐹 one and also a horizontal and a vertical component to the motion of the particle. That means we can write our total work π‘Š as the sum of the work in the horizontal and the vertical directions. We’ll call those values of work π‘Š sub π‘₯ for the horizontal work done and π‘Š sub 𝑦 for the vertical work done.

If we refer to our definition for work, we can write out equations for work in the π‘₯- and 𝑦-directions: π‘Š sub π‘₯ is equal to 𝐹 sub π‘₯ times 𝑑 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‘Š sub 𝑦, 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 𝐹 sub π‘₯ and 𝑑 sub π‘₯ and 𝐹 sub 𝑦 and 𝑑 sub 𝑦 from the given information in our problem. Take another look at 𝐹 sub one, the component in the 𝑖 or horizontal direction is 3.0 newtons. So we can write that in for 𝐹 sub π‘₯.

Then, as we look at the particle path, and specifically at 𝑁 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 𝑦 or vertical direction, we see that the component of 𝐹 sub one that is vertical is 4.0 newtons. And the displacement of our particle in the 𝑦-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‘Š sub 𝑦, 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‘Š, the net work done on the particle by 𝐹 one. To do this, we need to sum π‘Š sub π‘₯ and π‘Š sub 𝑦: 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 𝐹 sub one as the particle moves along this straight line path.

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