Work As Area Under F-x Graphs - Online Tutor, Practice Problems & Exam Prep

To calculate work done on an object using a force versus displacement graph, determine the area under the curve between the graph and the x-axis. This involves breaking the graph into simpler shapes like rectangles and triangles. The area of a rectangle is calculated as base times height, while the area of a triangle is given by 12 × ⁢base⁢height. Positive areas indicate positive work, while negative areas indicate negative work, leading to the total work done being the sum of these areas.

Calculating Work As Area Under F-x Graphs

So in this video, I'm going to show you how to calculate the work that's done on an object by using a force versus displacement graph. So some problems are just going to give you a diagram or graph like this that plots the force in the y-axis with the position in the x-axis. And they'll ask you to calculate how much work is done on an object by some force. We've actually seen something very similar to this in a previous chapter. So let's just get to it. The whole idea here is that the work that is done by any force, whether it's constant or variable, is really just the area that is under that f x graph. What that means, that phrase under the graph, is it just means the area that's between the graph and the x-axis. So really instead of using f d cosine theta a bunch of times to calculate work, all we do is we just figure out what's the area that is underneath the graph between where you are on the function and the x-axis. So there's an area right here. And then once it goes below the x-axis, there's also another area that's here.

So the whole idea here is that, again, instead of using f d cosine thetas, we're really just going to be breaking up this graph into a bunch of rectangles and triangles and using some pretty basic geometry equations to solve it. Let's check this out. So to calculate the work that's done by the force, we're just going to take the area that's underneath the curve. And what we can do here to make this a little bit easier for us is break this up into a bunch of simpler shapes. So if I break this up like this, I've got a rectangle like here, and I'm going to call this area one. So then I've got this triangle like this that's area 2. And then I've got another triangle that's going to be down here. I'm going to call this area 3. So the whole idea is that to calculate the whole work that's done on this object, I'm just going to be adding up all of those areas. Area 1, area 2, and area 3. So the area 1 is the rectangle. Area 2 is the triangle. And then area 3 is also a smaller triangle that's below the x-axis. Alright?

So let's get to this. Area 1. So to use area to figure out the area of a rectangle, we're just going to use base times height. So this is base times height. So I'm just going to look at this rectangle here and figure out the base and height by using the numbers on the axis. So I've got a base of 4 and then I've got a height of 30. So this is going to be 4 times 30, and I've got area 1 is equal to 120 joules. Let's move on to the second one. For area 2, we've got a triangle. So we're going to have to use the area for a triangle which is 12bh. So I've got 12bh. And if I go ahead and take a look here, the base of this triangle goes from 4 to 16, so that's 12. And then the height of this triangle is the same 30 that I was using before. So it's just going to be 12×12×30. And I get area 2 is equal to 180 joules. Alright? Now finally for the last area, area 3, that's the one that's sort of like underneath the x-axis over here. I'm going to use an area for a triangle, 12bh. This is going to be 12. Now I've got the base of this triangle is 4, and the height of this triangle now is not going to be 10 because we're actually going down into the negative axis here. So our height that we plug in is actually negative 10. So this is going to be 12×4×(-10), and what you end up getting is you end up getting negative 20 joules. So we can see here that when you have areas above the x-axis, those are going to be positive works because your force is positive. When you have areas below the x-axis, those forces are sorry, are going to be negative, and so you're going to have negative work. So, basically, we've got our 3 areas. Now we're just going to add them all up together and just, calculate. Right? So this is work done. This is going to be 120 plus 180 plus negative 20. So when you add all altogether, you're going to get the work that's done is equal to 280 joules, and that's your answer. So this is we can see here that this graph stuff is actually super straightforward. If you ever have a variable force, instead of using f d cosine thetas and trying to calculate like, you know, f averages or something like that, just try and see if you can create this sort of like f versus x graph. And so therefore, you can just calculate Anyways, that's it for this one, guys. Let me know if you have any questions.

Speed of a Remote Controlled Car Using F-x Graph

Guys, in this problem, we have a remote control car that is subject to a force that varies over some distance here. So, when we have these force-distance graphs, we're going to take the area under the curve, and that's going to be the work. We want to calculate the speed of the car at \(x = 4\). Now, the car initially starts from rest, so this is \(v_0 = 0\). We want to calculate the final speed. So, how do we do that? We're going to relate the work that's done to the change in kinetic energy by using the work-energy theorem. The net work is equal to the change in kinetic energy. I'm going to take the area under the curve for this whole entire graph here up until \(x = 4\), and then I'm going to relate that to \(K_{\text{final}} - K_{\text{initial}}\). I'm going to break this up into a couple of sections. The work done from 0 to 2, plus the work done from 2 to 3, plus the work done from 3 to 4. That is going to be equal to \(K_{\text{final}} - K_{\text{initial}}\). However, because the initial speed is equal to 0, there is no initial kinetic energy and all of this is going to be final kinetic energy, which then I can relate to the speed of the car. I'm going to take a look at the areas under the curves for each one of the three little terms that I've made. The work done from 0 to 2 is basically going to be all of this area right here. I'm going to highlight this in blue. That's going to be the sky. What about from 2 to 3? Well, from 2 to 3, there actually is no area under the curve because we're at 0. There is no force acting on this car from 2 to 3, so there is going to be no work done. Finally, from 3 to 4, there's definitely going to be some work done because it's going to be this area here, under the graph. I've got those highlighted areas here. Let's go ahead and start calculating.

So, if I'm going to calculate the work done from 0 to 2, I'm basically going to cut this up into 2 shapes: one triangle and one rectangle. I'm going to do this all in one sort of line, but this is going to come down to 2 terms. I've got the area of a triangle which is \( \frac{1}{2} \times \text{base} \times \text{height} \), which is going to be \( \frac{1}{2} \times 1 \times 20 \), and then plus the area over here, which is really just a rectangle. This is going to be base times height. So the base is 1 and the height is 20. So, if you go ahead and work this out, what you're going to get is you're going to get 30 joules. So, 30 joules is the first little section of this right here. So, actually, let me highlight this in blue. Let's go ahead and calculate the work done from 3 to 4. The work done from 3 to 4 is just going to be this little triangle right here. I've got \( \frac{1}{2} \times \text{base} \times \text{height} \), that's the base. And then the height of this thing, well, this doesn't hit all the way down to -20, and it's a little bit past -10. So, what I'm going to do is I'm going to call this -15 here. It's right in between. And remember, it's negative because we're in the negative y-axis here. So, the work that's done is actually going to be -7.5 joules, and that is the work done from 3 to 4. So, now, I really just add these two things together. And what this just becomes is I have 30 joules plus, this is -7.5 joules. And this equals \( \frac{1}{2} \times m \), so we know what the mass is. This is actually going to be 4 times \(v_{\text{final}}^2 \) here. So, this is \(v_{\text{final}}^2 \). I get 22.5 joules equals, this is going to be 2 \(v_{\text{final}}^2 \). So, you go ahead and work this out. What you're going to get is you're going to get 3.35 meters per second. So, that's the final answer here. That's the speed of the car at \(x = 4\). Let me know if you guys have any questions.

Here’s what students ask on this topic:

How do you calculate work done using a force vs. displacement graph?

To calculate work done using a force vs. displacement graph, determine the area under the curve between the graph and the x-axis. This involves breaking the graph into simpler shapes like rectangles and triangles. The area of a rectangle is calculated as $\mathrm{base}\times \mathrm{height}$, while the area of a triangle is given by $1/2\times \mathrm{base}\times \mathrm{height}$. Positive areas indicate positive work, while negative areas indicate negative work. The total work done is the sum of these areas.

What does the area under a force vs. displacement graph represent?

The area under a force vs. displacement graph represents the work done on an object by the force. This area is calculated by summing the areas of simpler geometric shapes (rectangles and triangles) that lie between the graph and the x-axis. Positive areas correspond to positive work, while negative areas correspond to negative work. The total work is the sum of these individual areas.

How do you handle negative areas in a force vs. displacement graph?

Negative areas in a force vs. displacement graph occur when the force is below the x-axis, indicating that the force is acting in the opposite direction of the displacement. To handle these, calculate the area as usual but assign a negative sign to it. When summing the total work, include these negative areas to account for the direction of the force.

What is the formula for the area of a triangle in a force vs. displacement graph?

The formula for the area of a triangle in a force vs. displacement graph is $1/2\times \mathrm{base}\times \mathrm{height}$. Here, the base is the length along the displacement axis, and the height is the force value. This formula helps in calculating the work done by the force over the displacement.

Why is it easier to use a force vs. displacement graph for variable forces?

Using a force vs. displacement graph for variable forces is easier because it allows you to visually break down the problem into simpler geometric shapes. Instead of using complex calculus or averaging methods, you can calculate the work done by finding the area under the curve. This method simplifies the process, especially when dealing with non-constant forces.