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A force 𝐹(𝑥) varies with position, as shown in the accompanying figure. Find the net work done by this force on a particle while the particle changes its position from 𝑥 = 1.0 m to 𝑥 = 5.0 m.
A force 𝐹 of 𝑥 varies with position, as shown in the accompanying figure. Find the net work done by this force on a particle while the particle changes its position from 𝑥 equals 1.0 meters meters to 𝑥 equals 5.0 meters.
As we look at this figure, we see that the force as a function of 𝑥 is on the vertical axis and the position of the particle, in meters, is on the horizontal axis. Along with the figure, we’re given two important pieces of information, that the particle changes its position from 𝑥 equals 1.0 meters to 𝑥 equals 5.0 meters.
In this problem, as we solve for the net work, we’re going to assume that this work is the only source of energy acting on the particle, that no other forces are at play.
Let’s begin by recalling the relationship between work, force, and distance. These three qualities are related this way. Work is defined as the force exerted on a particle multiplied by its displacement; that’s how much work is done on a particle.
Now in this problem, we wanna solve for the net work done. We’ll call it 𝑊 sub net. And this work will be equal to the force exerted on our particle multiplied by the displacement of the particle.
To begin figuring out the net work done on the particle, let’s draw an expanded view of this force versus position diagram. So here is our expanded view of that same figure we saw earlier. It’s the force on the particle related to the position of the particle.
And remember, we wanna solve for the network on the particle when it goes from the position of one meter to five meters, so let’s mark those two spots on our plot. So here, we have the start and stop points of our particle, 𝑥 equals 1.0 and 𝑥 equals 5.0 meters, and we want to calculate the net work done on the particle between these two points.
Now if we had an example of a constant force over some distance, then we would simply multiply that constant force by the displacement of the particle over which that constant force acted to find the work. But in this case, we have a variable force over our displacement. To find the work then, we’ll need to find the area under the curve of our force versus position curve.
When we find that area and add all of it up over the course of our journey from 𝑥 equals 1.0 to 𝑥 equals 5.0 meters, then we will have arrived at the net work done on a particle. As a preview of what’s to come, we’re gonna find the area under this curve to be both positive in this region — so there is positive work there — and negative when the curve goes below the 𝑥-axis in this region there between four and 5.0 meters.
So what are these areas? We’ll solve for the area of positive work, the area where force and position multiply together to give a positive number first and then we’ll solve for the amount of work that’s negative, combine the two, and that will give us the net work done on our particle.
To find the area where positive work happens, let’s break up this shape above the positive 𝑥-axis into a rectangle and a triangle, like this. Now we can solve for the area of a rectangle and the area of a triangle and add them together to give the total positive work done on this particle.
Looking carefully at this diagram, we see that our rectangle has a width of three meters, and it has a height of one newton. When we multiply these numbers together, we find the area of a rectangle to be three newton-meters.
We’ll make a note of that off to the side where we’ll say 𝐴 sub 𝑟, the area of the rectangle, equals three newton-meters. Now we move on to the area of our triangle which we’ll call 𝐴 sub 𝑡. Now we’ll recall that the formula for the area of a triangle is that the area equals one-half the base times the height of the triangle.
Now with this triangle we’re considering, the base we just solved for is three meters, and the height it turns out is four newtons. So four newtons times three meters times one-half is equal to six newton-meters.
So now we’ve solved for the area of our triangle and the area of our rectangle, which are both positive work values. To find the total positive work done, we add these together and find a result of nine newton-meters. And we’ll call this value 𝑊 sub plus because it’s the positive work done on the particle.
So now that we found the positive work done on our particle during its journey, let’s solve for the negative work done on the particle. To do that, we’ll focus on the section of the particle’s path between four and 5.0 meters. Once again, we have a triangle whose area we want to find.
The width of this triangle is one meter, and its height is two newtons. When we refer back to our area of triangle equation, we see that the total area of this triangle is one meter times two newtons times one-half or one newton-meter.
Now because this is work that is negative, meaning that the force exerted over this distance is a negative force, we’ll call this 𝑊 sub negative or 𝑊 some minus. And that equals negative one newton-meter.
We’ve now solved for both the positive and negative parts of the net work done on our particle during its journey. So to find the net work done between 𝑥 equals 1.0 and 𝑥 equals 5.0 meters, we’ll combine the positive and negative work done.
The network is equal to 𝑊 sub plus plus 𝑊 sub minus, or nine newton-meters minus one newton-meter, which equals 8.0 newton-meters.
We use two significant figures because we were given two significant figures in the position of our particle. Now another way to write a newton-meter is in a condensed unit called joules, so we can also write the net work done on our particle as 8.0 joules represented by a capital 𝐽.
And there we have it! The net work done on our particle as it moves from 𝑥 equals 1.0 to 𝑥 equals 5.0 meters in position.
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