Work On Inclined Planes - Online Tutor, Practice Problems & Exam Prep

To calculate the work done by gravity on an object on an inclined plane, break down the gravitational force (mg) into components: mgx (parallel to the incline) and mgy (perpendicular). Use the equation for work, W = Fd × cosθ, where the angle is between the force and displacement. For mgx, the work is 7,080 joules, while mgy does no work (0 joules) since it acts perpendicular to the motion. Thus, the total work done by gravity equals the work done by mgx.

Work By Gravity On Inclined Planes

Guys, in this video, I want to show you a specific example of calculating the work that's done by gravity, but in a specific case where you have an object on an inclined plane. The reason I'm doing this is because these problems can be kind of confusing and tricky. They sometimes try to trick you into plugging the wrong number into your equations. So I want to go over this so to make sure that you actually don't make that mistake. So we're actually going to start with the example and we'll come back to this point in just a second here. In this problem, we have a 100 kilogram block that's on an incline, and it's going to slide down a distance of 12 meters. We also know that the angle of the incline is 37 degrees. What we want to do in this problem is we want to calculate the works done by mg_x, mg_y, and mg. So, basically, the works done by each one of these, the components of mg and then also mg itself.

So how do I do that? Well, I want to go ahead and draw those forces in my diagram. Right? In the inclined plane, my mg points straight down. My normal points perpendicular to the surface. And because we tilt our coordinate system, then we just have to break up this mg into its components. So this is going to be my mg_x and this is going to be my mg_y. So I'm trying to figure out the works done by the components of mg and then mg itself. Alright? So how do I figure out works? I'm always just going to use fdcosine theta. So we're just going to use mg_x × d × cosine(θ), and we're going to use mg_y × d × cosine(θ), and then mg × d × cosine(θ).

What I want you to remember is that the θ term that we use in our fdcosin(θ) always refers to the angle that is between the force and the displacement or the distance. Basically, it's the angle between the force and the motion. That's why I always draw an arrow between this θ and my f and my d just so I don't forget that. Okay. So what I have to do is I want to figure out mg_x × d × cosine between these two vectors here. So let's get started. How do I figure out mg_x in an inclined plane? Well, remember, mg_x is going to be mg × the sine of the incline angle. This 37 degrees here is actually θ_x. It's the angle with respect to the horizontal. That's my 37 degrees there. So really what happens is I can replace this mg_x with mg × sin(θ_x) here. So we're doing mg × sin(θ_x) × d × the cosine of θ here. So here's what I want to point out, is that these problems, with inclined planes, be careful not to plug in the incline angle, this θ_x here. Don't plug in this θ_x here for cosine(θ), which is really in your fdcosine(θ). Right? So don't get these two angles confused. This is the angle of the incline, your θ_x. This is the angle that is between these two forces. They're not the same thing.

Alright? So let's go ahead and do this. Right? We can actually go ahead and plug in all of our numbers. We have m as 100, we have 9.8 for g. And then we have the sine of θ_x. That's my 37 degrees here. Now I multiply it by the displacement or the distance, and that's 12 meters down the ramp. Now the cosine is going to be the cosine of the angle between your mg_x and your displacement. If you take a look at your diagram here, mg_x points down the ramp, and your displacement also points down the ramp, which means that the angle that we plug in here is actually 0, not 37 degrees. And remember that the cosine of 0 always evaluates to 1. So, once you plug all this stuff into your calculator, you get 7,080 joules. So that's the work done by mg_x. Now let's take a look at mg_y. Basically, we're going to do the exact same thing here except now we're going to do mg × the cosine of θ_x, right because that's mg_y × d × cosine(θ). And we plug this we're just going to plug in all the numbers. This is going to be 100 times 9.8 times the cosine of 37. Now we have times 12. And now we just figure out the cosine of the angle between these two forces, right between my mg × cosine(θ) or my mg_y and the distance. What is that angle? Well, if you take a look at your diagram, your mg_y points into the incline and your displacement points down the incline. Those are perpendicular to each other. Right? This is actually a right angle like this. So what happens here is we actually plug 90 into our cosine term, not 37 again. And remember that the cosine of 90 always evaluates to 0. So even though you have a bunch of numbers out in front of here, you're going to multiply it all by 0. And the work that's done by mg_y is always just going to be equal to 0 joules. So this is actually always going to be the case. And you can think about it like this. Right? Your mg_y always points into the incline, but you're always moving down the incline so it can never do any work like that.

The last thing we do is mg × d × cosine(θ) here. So what we have to do now is actually look at this force, which is our mg, and we have to figure out the angle between this mg and this displacement here. And this is where it gets pretty complicated because this angle here in our inclined plane problems is actually θ_y. It's the angle with respect to the y-axis which we're almost never given. So this actually ends up being really complicated here and we're not going to use this approach to solving the work done by gravity. Fortunately, we can add them up just like regular numbers. So if I want to calculate the work done by gravity, it's really just the addition of the work done by mg_x and the work done by mg_y. I can add these two works together even though they come from forces that point in the different directions like x and y because works are actually scalars, they're not vectors. And what this allows us to do is our work done by mg_y, remember, is always just equal to 0, so we can just cross that out. So, really, what happens here is that the work done by gravity is always just the work done by mg_x. And this should make some sense. So this is going to write W_mg is always equal to the work done by mg_x. And the reason for that is if you think about this, the component of gravity that pulls the block down the ramp is mg_x. That's the only thing that can do work on the box. So what happens here is that our work done by mg is just 7,080 joules, just like your mg was. And that's the answer. So you have 7,080 joules and so your work done is equal to mg_x. By the way, this is also always going to be the case.

Alright. So that's it for this one, guys. Hopefully, that made sense. And let me know if you have any questions.

Pulling Crate Up An Inclined Plane

Guys, in this problem, we are pulling a crate up an inclined plane, so I am just going to go ahead and draw that out real quick. About my inclined plane like this, I've got my box, and I know that the mass is equal to 19 kilograms, so m equals 19. Now I am going to go ahead and label all the forces that are acting on this crate. I have an applied force that is parallel to the ramp and it is 130 newtons. So this f_a here is a 130. Now, I also have some gravity. Right? So I have this m g here. This is going to be my mg, and I have a normal force like this. Now remember that on our inclined planes, we are going to separate our mg, and we are going to have an m_gx that points down the incline like this. And we actually are told nothing about friction, so we are not going to assume there is any friction in this problem. So in this first part here, I want to calculate the work that is done by mg. Well, remember that on an inclined plane, the work that is done by mg is the same exact thing as the work that is done by m_gx. This just allows us to simplify our equation. And really, this is just going to be mg times the sine of the theta x, in which theta x is the angle of the incline. Here, we are told this is 36 degrees. So that is the angle we are going to plug into this theta x here. So it is mg sin theta x times d, and we have already taken care of that cosine theta term. It is just this right here. Now, what we have to do is we have to determine a direction for positive because that is going to influence whether our works are positive or negative. Now because we are pulling this 19 crate up the ramp, so that means that the displacement over here is actually going to be this way. So this distance here, we are told is 15 meters, then I am going to choose this direction to be positive. And that means that this work done by gravity is actually going to be negative. Alright? So this is just going to be negative, and now I am just going to plug in all my numbers. So I have got the mass which is 19 times the 9.8 times the sine of theta, which is the sine of 36 times the distance that I am pulling it over, which is 15. Now again, there is no need to put this cosine theta term here, this cosine between this m_gx and d because the negative sign already takes care of that. And remember that m_gx is going to be parallel to the displacement, so you actually do not need to do this. Alright. So you just plug in all these numbers here, what you are going to get is negative 1640 joules. So that is the answer to part a. Alright. It makes sense that it is negative because as you are going up the ramp, gravity is going to be doing negative work on you. Alright. So let us go ahead and move on to part b. In part b, we want to figure out the final kinetic energy of the crates. We just calculated a work. How do we relate this to kinetic energy? Remember, we just use the work energy theorem. The net work on an object, which is just the sum of all the works, is just going to equal to the change in the kinetic energy. Now, in this case, remember we have 3 forces. We have our applied force, our m_gx, and the normal force. So when we go to sum all of our works, the net work is going to be the work done by the applied force, the work done by friction, which we just calculated. Oh, I'm sorry. That the work done by friction, the work done by gravity, which we just calculated, plus the work done by the normal force. Now remember the normal force is sort of perpendicular. Right? It is going to be pointing, sort of in the new y axis, perpendicular to the incline, and it is going to be perpendicular to the direction of motion. So there is no work done by the normal force. So it is just these two that contribute works. And that is equal to just change in the kinetic energy. That is kinetic energy final minus kinetic energy initial. We know that we are starting from rest. Our v₀ equals 0, so there is no initial kinetic energy. So really the net work, all these works here are going to be really just do are are going to equal the kinetic energy final. So I have already just calculated the W_mg in part a. Now all I have to do here is I just have to calculate the work done with the applied force, and then I can figure out the kinetic energy final. Remember that the applied force is just going to be f_a times d cosine theta. So that means the work done by the applied force here is going to be my applied force of a 130 times the distance is 15. And then what about the angle? Well, the f_a and the distance are going to be pointing in the same direction. They are going to be parallel, so this is just going to turn to the cosine of 0 which is just 1. And so if you plug this in, what you are going to get is positive 1950 joules. So now this is a 1,950 joules. It is positive. It makes sense because it points in the direction of motion. Now we are just going to plug that back into our FA. So really the work nets, which is really just the w_fa, which is my 1950, plus my work done by gravity, which is negative 1640, is going to equal the final kinetic energy. So basically, what happens is that the kinetic energy final is going to be equal to 310 joules, and that is your answer. Alright. So that is it for this one, guys. Let me know if you have any questions.

Finding Speed of a Crate Using Work-Energy Theorem

Guys. So in this problem, I have a crate that's sliding down an inclined plane and it has some forces acting on it, and we have to calculate some works. So let's go ahead and just draw a quick little sketch here. I've got this inclined plane like this. I'm going to draw it nice and big here. I've got my box that is 7 kilograms, and I've got some forces acting on it. So let's go ahead and label all the forces. I've got m g, I know that's acting. I've got the normal force. We've got a component of m g that's acting down the ramp. This is m g x, and this is my mg y. And then I also have the force of friction. As the block slides down the ramp, friction wants to oppose that and wants to go up the ramp like this. So I got my friction force. So now what I want to do is calculate the work that's done by gravity if the crate slides 2.5 meters down the ramp. So basically, that's this distance like this, d equals 2.5 here in which it's going to slide down the ramp. So it's going to end up over here somewhere. So let's go ahead and do that in part a. In part a, I want to calculate the work that is done by gravity. So it's W_MG_. Now remember that on an inclined plane, W_MG_ is the same thing as W_MG__x. And so I can just say this is going to be m g times the sine of theta x. This angle here is 26 degrees. So this is theta x, equals 26 degrees. And, it's going to be m g sine theta x times the distance. So what I have to do is first I have to pick a direction of positive. Now what happens is my m_g_x is going to actually point in the direction of my displacement. And so what I'm going to do, I'm just going to have my positive direction be this way. So what this means here is that your W_m_g is just going to be, this is going to be positive and so therefore, I'm just going to start plugging stuff in. So I've got 7 times 9.8 times the sine of 26 degrees and then I've got the distance here which is 2.5. If you go ahead and work this out, what you're going to get is you're going to get 75.2 joules. Makes sense that this is positive because the blocks are going to start going faster because of this work here. Alright. So what about part b? In part b, now I want to calculate the work that's done by friction. So remember that when we calculate the work done by friction, it's always going to be negative f_k d. So this is just going to be, once I expand this out, negative mu_kinetic times the normal times distance. So really, the trouble is that I want you to figure out this normal force on an inclined plane. Now, on an inclined plane, if your only if your only forces act horizontally, then your normal force is going to equal m g y, which is just going to be m g times the cosine of theta x. So what I can do here is I can see that the work that's done by friction is going to be negative mu_k, and then the normal force really just becomes m_g_cosθ x, and then I have d here. So I look through all my variables. I have mu_k, I have the mass, gravity, I have the angle, and the distance, so I can calculate the work done by friction. So my work done by friction is going to be negative, 0.36 times 9.8 times oh, I'm sorry. The mass, the mass is 7. The mass is 7. Now we have 9.8. Now we have the cosine of the angle which is 26 degrees and this times the distance which is 2.5. So if you go ahead and work this out, what you're going to get is 55.5, except it's going to be negative. So it's negative 55.5 joules. Makes sense that it's negative because as the block goes down, grab so the first force of friction is going to want to remove energy from the block and slow it down. So it's negative. Alright. So now in this last part here, we have to find the speed of the crate if it starts from rest. So the idea here is that the initial velocity is going to be 0 here, and then when it gets down to this point, it's going to have some final velocity, and I want to figure out what that is. So I just calculated a bunch of works, and I can relate them to speeds by using the work-energy theorem. Remember that the net work is just equal to the change in the kinetic energy. So the work that we just calculated, there are the works that we just calculated, 75.2 and negative 55.5, we can combine them to find the network. So our network is really just going to be our m_g_, plus the work done by friction, plus the work done by normal. Now in this case, the normal doesn't do any work. Right? We know that. So this is equal to the change in the kinetic energy. So this is really just k final minus k initial. However, because the initial speed is equal to 0, there is no initial kinetic energy and all of it just goes to the kinetic energy final. Alright? So that's all we gotta do. So really, we just have this 75.2 that we just plug in plus negative 55.5, and this equals the kinetic energy final. This is going to be 1/2 m v_final². We want to find this v_final here. We're just going to have to solve everything else. So again when you subtract these two things is you get 19.7. This equals 1/2 and then times 7 times v_final². Let me rewrite that again. So this is 7 here. So when I move everything over to the other side, what I'm going to get is 19.7 times 2 divided by the mass of 7, and this equals the v_final². So when you go ahead and plug this in and take the square roots, what you're going to get is that your v_final is equal to 2.4 meters per second, and that's your final answer. Alright, guys. So that's it for this one. Let me know if you have any questions.

Here’s what students ask on this topic:

How do you calculate the work done by gravity on an inclined plane?

To calculate the work done by gravity on an inclined plane, you need to break down the gravitational force (mg) into two components: mg_x (parallel to the incline) and mg_y (perpendicular to the incline). The work done by gravity is primarily due to the mg_x component. Use the equation for work, $W=Fd\times cos\theta$, where θ is the angle between the force and displacement. For mg_x, the work is calculated as $W=\mathrm{mg}\times sin\theta \times d\times cos0$. Since cos(0) = 1, the work done by mg_x is $W=\mathrm{mg}\times sin\theta \times d$. The mg_y component does no work as it is perpendicular to the motion.

What is the role of the angle in calculating work on an inclined plane?

The angle plays a crucial role in calculating work on an inclined plane. In the equation $W=Fd\times cos\theta$, θ is the angle between the force and the displacement. For the component of gravity parallel to the incline (mg_x), θ is 0 degrees because the force and displacement are in the same direction. For the perpendicular component (mg_y), θ is 90 degrees, making the work done by mg_y zero since cos(90) = 0. It's important not to confuse the incline angle with the angle between force and displacement.

Why does the perpendicular component of gravity (mg_y) do no work on an inclined plane?

The perpendicular component of gravity (mg_y) does no work on an inclined plane because it acts perpendicular to the direction of motion. Work is defined as $W=Fd\times cos\theta$, where θ is the angle between the force and displacement. For mg_y, this angle is 90 degrees, and cos(90) = 0. Therefore, the work done by mg_y is zero. This makes sense because mg_y does not contribute to the motion along the incline.

How do you determine the angle between force and displacement on an inclined plane?

To determine the angle between force and displacement on an inclined plane, you need to analyze the direction of each component of the gravitational force relative to the motion. For the parallel component (mg_x), the force and displacement are in the same direction, so the angle θ is 0 degrees. For the perpendicular component (mg_y), the force is perpendicular to the displacement, making the angle θ 90 degrees. Always ensure you are using the angle between the force vector and the direction of motion, not the incline angle itself.

What is the total work done by gravity on an inclined plane?

The total work done by gravity on an inclined plane is the sum of the work done by its components, mg_x and mg_y. Since mg_y does no work (0 joules) because it is perpendicular to the motion, the total work done by gravity is equal to the work done by mg_x. This can be calculated using $W=\mathrm{mg}\times sin\theta \times d$. For example, if a 100 kg block slides down a 37-degree incline for 12 meters, the work done by gravity is 7,080 joules.