Intro to Calculating Work - Online Tutor, Practice Problems & Exam Prep

Work (W) is the transfer of energy, calculated using the equation W=Fd⁢cos(θ), where F is the force, d is the displacement, and θ is the angle between them. Positive work occurs when the force aids motion, while negative work occurs when it opposes motion. For gravity, work is positive when falling and negative when rising, illustrating energy transfer in kinetic energy (KE) as KE = 1/2 × mv^2.

Work Done by a Constant Force

Guys, so now that we understand kinetic energy, in this video, I want to start talking about work. So we're going to see the relationship between work and energy and we're also going to see the relationship between work and forces. I'm going to show you what work is and then we'll see how to calculate the work done by a constant force. Let's check this out.

First of all, what is work? Well, imagine that we had a box at rest on a frictionless surface. Right? So the velocity is equal to 0. So if we wanted kinetic energy, the kinetic energy is also equal to 0, and then you start pushing this box. So if you push it and it's frictionless, then the box is going to start to move. It's going to start to move to the right which means that now it has some velocity that's not equal to 0 and so, therefore, it has some kinetic energy that's not equal to 0 as well. So if the box gains some speed by you pushing it some velocity, it also gains some kinetic energy. Right? That kinetic energy is always associated with your speed or your motion. So where did that energy actually come from? And the whole idea is that the energy actually comes from you. It comes from the force that you were pushing on this box with. So remember that energies in physics are always transferring between objects. The sun gives energy to the plants and then a cow eats the plant, that's a transfer of energy, then you eat the cow, that's a transfer of energy, and then you push the box. So the whole idea here is that work, which is written by the symbol w is actually the quantity. It is the amount of energy that you transfer between objects or that is transferred between objects. So you push the box and let's say it gains 10 joules of kinetic energy, you do work on the box. That's what we say in physics. You do work on the box or objects do work on others. So because work is really just energy that's transferred, then the unit that we'll use for this is going to be the joule just like, you know, just like energy.

Alright? So let me show you the examples for work. How do we calculate? Or so the equations for work. How do we calculate it? And I'm going to show you the fancy version of this that you'll see in your textbooks. It's the magnitude of the force times the magnitude of the displacement times cosine θ. So a lot of textbooks will rewrite this as a shortcut way, and this is how we're going to write it from now on, fdcosθ. So really, what we have to know about these equations, about this equation in particular, angle. θ is always the angle between f and Δx or f and d. So, basically, in these problems where you calculate work, you're going to have to know 3 things. You're going to have to know the magnitudes of both the force and the displacement. And remember, the magnitudes are always going to be plugged in as positive numbers. And then you're going to figure out the angle that is between those vectors. And that's what you plug into your cosine θ term.

Let's check out some examples. So here, we're going to pull a 2 kilogram box that's at rest on a frictionless surface. So the idea is this is 2, and then I'm going to pull this box with a force of 3 Newtons. And I want to calculate the work if I pull it through a distance of 5 meters. So basically, what happens is I want to figure out the work that's done by my force. So remember that the work is going to be fdcosθ, and I'm going to plug in everything as positive. So I have the force, which is 3 newtons. And if I'm pulling it to the right, then, therefore, it goes through a displacement, a Δx of 5. So really what I what happens is I'm going to plug this in and then I'm going to plug in 3 times 5. And now what's the cosine of this angle between those two vectors? Well, if these two vectors are parallel to each other, they point in the same direction, then therefore, the cosine of this angle here, this angle is equal to 0. And so with the cosine of 0 is, it's always equal to just 1. So, really, all we do is we actually just multiply 3 times 5 and we get 15 joules.

Alright. So that's the work that's done. Let's move on to the second example. Now we have a cart, a 5 kilogram cart here, that's already moving to the right with some velocity. So the idea here is now I have this cart like this. Now this is 5. The velocity of this cart, the initial, is already equal to 10 meters per second, but there's going to be some stopping force. I'm going to call this fs, and it's a 100 Newtons. Now I want to figure out the work that's done by the stopping force. I'm just going to use the same equation, fdcosθ. So this is my work done is going to be fd, and this is going to be the stopping force times d times the cosine of θ. Alright? So I've got the stopping force. This is 100. Now what about the distance or the displacement? Well, if the cart is already moving to the right, but it's being pulled to the left, it still undergoes a displacement that's to the left here. This Δx is 2.5. So this is a 100 times 2.5. And now, what's the angle that is between these two vectors? Well, we said that 0 degrees was when they're parallel to each other. But when they're antiparallel, when you have a force like this and a displacement like this, then the angle is actually equal to 180 degrees. Here, it was equal to 0. So what happens here is that when you plug in cosine or θ equals a 180, your cosine term will always turn to negative one for antiparallel forces. Right? So, really, this just turns into a 100 times 2.5 times negative one, and this is just 250 joules, but it is negative.

Alright? So we can see here is that work can actually be positive or they can be negative. Remember that work is really just the amount of energy transfer. Work is the quantity of energy that any force can either give to or take away from an object. So work can either be plus or minus. And the simple rule to figure this out is to look at the force and figure out whether it helps or hurts the motion. If the force helps, meaning it goes along with the force or the object's motion, then the work that's done is going to be positive, like we had over here. The force causes the cart to move or the box to move to the right. The force goes along with the motion. The work is positive. Now if the force hurts or goes against the object's motion, the work is going to be negative. That's basically what we had in this situation here. The cart was already moving to the right. This is the motion, but the force points this way. It hurts or takes away from that kinetic energy, and that's why we got a negative number.

Alright? So that's it for this one, guys. Let's move on.

You pull a 5kg box vertically up with a constant 100N force for 2m. How much work do you do?

• A. 51 J
• B. 100 J
• C. 102 J
• D. 200 J
• E. 500 J
• F. 1000 J

Work by a 2D Push/Pull

Guys, let's check this one out here. So we've got this 10 kilogram suitcase, that's the mass of the box. And we're also pulling with some applied force of 50 newtons. Now we're pulling it across the floor, and it's going to be pulled some distance of 20 meters. But which way is it going to be pulled? Well, even though we're pulling it upwards like this, what happens is it's just going to move along the floor. Right? It actually wouldn't fly up like this. So what's going to happen is it's going to travel some displacement here. Let me draw this a little bit bigger and we know that this delta x, or d, it doesn't matter which letter we use, is going to be 20 meters. So we want to do this first part here is we want to figure out the work that's done by the pull. Right? So your pull here so the work that's done by f. Well, remember, the work that's done by any force is always going to be fd⁢cos⁢θ. So the work done by f is going to be fd⁢cos⁢θ. Remember that this angle is the angle between f and d. So let's check this out. I have the magnitude of the force. It's 50. I have my x or my d. Right? It's just equal to 20 meters. So I can just plug this into my work equation. This is going to be 50 times 1 sorry. 20 times the cosine of the angle that's between these two vectors. We can take a look here. They're actually not parallel. Right? The force actually points this way, but your displacement points along the floor like this. So this is the angle that we use between these vectors, and we know that this angle here from the problem is equal to 37 degrees. So we're just going to plug this in, 50 times 20 times the cosine of 37. You go go ahead and work this out. What you're going to get is you're going to get 800 joules. So that's the answer to part a. Alright. So now in part b, what we have to do is we're going to have to decompose our force into its components, f_x and f_y, and then calculate the works done by each one of those components. So for part b, we have the work that's done by f_x. Remember what we said, the work done by any force is that force, so f_x times d cosine of theta. Now we actually don't have f_x in our diagram just yet. So what we're going to do is we're going to take this force and we're going to break it down into its components like this. So this is going to be my f_x and I know to figure that out, I'm just going to use f times cosine of theta here. So this is just going to be 50 times the cosine of Actually, let me go ahead and write it somewhere else. So I've got that f_x is just equal f cosine theta. So it's just equal to 50 times the cosine of 37. So if you go ahead and plug this in, you're going to get 40. And the same thing is going to happen with f_y except now we're just going to use sine, right? So f_x is f_y is f sine theta. That's basically this vector like this. This is my f_y. And this is going to be 50 times the sine of 37. So you're going to get 30 here. So those are the components. It's just basically a 30-40-50 triangle, which makes sense. So now what we have to do is we have to calculate our work. So we're going to take the magnitude of f_x, which we know. This f_x is really just equal to 40, remember? So this is going to be my 40. Now the distance that it travels is still going to be the 20. That doesn't change. But now what's the cosine between the between or what's the angle between these two vectors? Well, now what happens is your f_x, remember, actually points along the horizontal in the same direction as your displacement. So what's going to happen here is you're going to have 40 times 20 and the cosine between these two angles is actually going to be or sorry, between these two vectors is actually 0. Right? Because basically your f_x points to the right and your distance also points to the right here. So what happens is you get a cosine of 0 and then we know that equals 1. If you plug this in, you're going to get 800 joules. So we get the same exact number but and that's basically because the only force that's actually doing work on this object is this f_x here. That's what gives you the 800 joules. One way you could also think about this is that we use fd⁢cos⁢θ when we calculated part a. And in part b, what happens is you already kind of have a cosine of theta that is absorbed in this term here. It's kind of wrapped up inside this f_x because really, this is already f times the cosine of theta. So you're doing fd⁢cos⁢θ here, but you're doing f times cosine theta times d when you're doing it this way. So you're just going to get the same exact number here. Alright. So to finish off the problem, we're going to take a look at part c. This is going to be the work that's done by f_y. So this is going to be my f_y times d cosine of theta. So we take a look. We got f_y. Remember this? That's just the 30 that we calculated over here. Times the displacement which is 20. But now what's the cosine or what's the angle between these two vectors? Well, basically, with this, f_y is vertical, but we know our displacement actually points to the right like this. So the angle between these two is actually 90, and so, therefore, the cosine of 90 is going to be 0, and the work that's done is just basically going to cancel out to 0 joules. That makes sense because there's a component of your force that's pulling vertically, but if you're traveling horizontally, then there's no work that that force does. Alright, so that's it for this one, guys. Let me know if you have any questions.

Work Done By Friction

Guys, so in this problem, we're going to be calculating the work done by a bunch of these forces: kinetic friction, weight, and normal. So let's go ahead and check out this problem here. We have the box that weighs 50 newtons, which means that your mg, which goes down, is already equal to 50 newtons. Don't assume that to mean 50 kilograms, and then you multiply by 9.8, because you're going to get the wrong answer. The mg is already 50, which means that the normal force, right, these are the only two forces that are acting in the vertical direction, have to cancel, so that's our normal force. We actually have one last force to consider, which is the force of friction. So we're pulling this block to the right, and friction's going to oppose that by acting to the left here. So this is going to be our F_k. That's the friction force. Now, what we want to do in this first part here is we want to calculate the work that is done by this kinetic friction. So, that's the work that's going to be done by F_k. Well, remember, if we're trying to figure out the work done by F_k, that's going to be F_k × d × cos(θ), so that's going to be the equation for that. We have this θ here. That's the angle between F_k and d. Alright. So how do we figure out this force of friction? Well, remember the force of friction is always going to be μ_k × the normal force. So we have μ_k × normal × d × cos(θ). Alright. So we have this coefficient of friction here. This is μ_k is 0.7. Now we just have to figure out the normal force. Well, actually, we already know that. It's equal to 50. So we're ready to start plugging in. The work that's done by the friction force is going to be 0.7 × 50. That's the normal force. Oops. We got 50 over here. Now we got the distance. Right? So the box actually travels a horizontal distance of 8 meters. So basically, we've got this distance over here, this Δx or d is going to be 8. So that means that we're just going to plug in 8 into here, and now we're just going to look at the cosine of the angle. Remember, the cosine of the angle or the angles between the force of friction and your displacement. Your friction force acts to the left, but your displacement acts to the right. So here, what happens is your force of friction or your F_k is this way, but your displacement is this way. This is your Δx or your d. And so what that means here is that the angle is equal to 180 degrees. So your cosine is going to be the cosine of 180. Now remember what happens is that when you plug in the cosine of 180, you're going to get a negative one always. So what this means is that you're going to get the friction force is equal to negative 280 joules. Now, what I want to point out here is that the friction force is always going to be negative. Your friction force is always going to oppose your direction of motion. We were going to the right in this example, and our friction force pointed to the left. If it was reversed, if we were going to the left, our friction force would actually point to the right. So because friction always opposes your motion, it's always in the opposite direction as your v, it always does negative work. So what you're always going to see when you calculate the work done by friction, is you're going to see f d cosine, and the number here is always going to be 180 degrees. They're always going to be opposite of each other. So one way we can simplify this from now on is we'll write that the work that's done by friction is just negative F_k × d. Alright? That's a pretty simple way to figure that out. So you plug in those numbers, and you'll get negative 280 joules. Alright. So let's move on to part b, now. Now we're going to calculate the work done by the weight force. So the work done by weight is really going to be the work done by gravity, work done by mg. So here, we're just going to do mg, that's the weight force, × d × cos(θ). But we can quickly tell here that your mg is going to point downwards, and your displacement is going to point to the right. So this angle here is 90 degrees. So what happens here is you're going to have MG, which is 50 × the displacement, which is 8, but it doesn't matter because the cosine of 90 is going to wipe everything out, and the whole entire thing is going to become 0. So what happens is the work that's done by mg is going to be 0 joules. That makes sense. Your weight force always points down. If you're going to the right, that weight force doesn't actually do any work on you. Alright? Now I got one last part here, which is the work done by the normal force. So this is the work done by this N. So this is going to be our normal force × d × cos(θ), and what we're going to see is that this basically is going to be exactly the same as part b. Your normal force points up, and your displacement is to the right, so this angle here is 90 degrees. So you have your weight or sorry, your normal is 50. Your displacement is 8, but it doesn't matter because you're going to have the cosine of 90 again, and so this whole thing is going to go away. You're going to have the work of the sun as 0 joules. Alright. So that's it for this one, guys. Let me know if you have any questions.

Work Done By Gravity

Guys, now that we understand how to calculate work done by constant forces, in this video, we're going to take a look at a specific force, which is the force of gravity. I'm going to show you how to calculate the work that is done by gravity. The whole idea here is that because gravity is a force, the m*g that we always plug into our free body diagrams, then that means that gravity can do work. Just remember, a quick definition of work is that it's the transfer of energy. And the way we figure out whether it's positive or negative is it's positive when the force goes along with the motion, when it helps your motion, and it's negative work whenever the force points against your motion. Right? So, we're really just going to take a look at 2 different scenarios here. Gravity acts vertically. Right? Your m*g points down. So we're going to look at two different situations when objects are falling, going down, or when they're rising and going up. Let's just get to the example here. Right?

In this first example, we have a 5.1-kilogram book that's falling from a 2-meter bookshelf. We've got this diagram here to help us out with this. We've got some information about the speed right before it hits the ground. In this first part here, we want to calculate the work that is done on the book by gravity. We want to figure out Wg. Remember, when we calculate work, you're always just going to start from Fdcosθ. You need to know the force, the distance, and the angle between those vectors. Our force F really just becomes mg. So your distance now, if your book is falling downwards, then it is just going to undergo displacement of Δy, and it's going to go downwards like this. Right? The angle between those two vectors is just 0 degrees. θ = 0, and we know that this term just becomes 1. So your work is really just W = mg × Δy. We've got all of our numbers right; we've got the mass which is 5.1, we've got 9.8 for g, and then the displacement, the Δy, is really just the 2 meters that it falls. You're going to get a work that is 100 joules, and that's the first part.

Part B asks us to calculate the kinetic energy right before it hits the ground. This is going to be the kinetic energy final, really. When the book is on the bookshelf, our kinetic energy is actually equal to 0. Right? It's not moving. When gravity is going to pull this thing downwards, it's going to have some velocity right before it hits the ground. Remember that kinetic energy is really just 12mv2. Kinetic energy is just going to be 12 times 5.1 times 6.262, and you go ahead and plug this in, and you're going to get a kinetic energy of 100 joules. Notice how we get the same exact number here, and that is no coincidence. Remember that work is really just a transfer of energy. When the force of gravity does 100 joules of work on the box, then the box basically gains 100 joules of kinetic energy.

Alright, let's move on to the second part now, which is when objects are rising or going up. We're going to take a look at a problem here in which a rock is thrown vertically upwards, and we want to calculate the work and energy. The setup of these problems is going to be the same. Your gravitational force still points downwards. But now because this block has a velocity that's up, your displacement vector is actually going to point up now instead of down. So, really, when you go through your Fdcosθ, what happens is that your force is still going to be mg. Your distance is still going to be Δy. Right? But now, the angle between those two is going to be 180 degrees because your m*g points down, and your Δy points up. So your cosine is going to be -1. So your work that is done by gravity just becomes negative mg × Δy. When you have objects that are going up, it's just going to be negative. This is going to be 2 for your mass, 9.8, and the maximum height is 11.5. So you're going to get the work of negative 225 joules.

You push a 3kg box against a wall for a distance of 2m with a force of 40N that makes a 53° angle with the horizontal, as shown. Calculate the work done by gravity.

• A. 58.8 J
• B. -58.8 J
• C. - 160 J
• D. -80 J

Work Done on Curvy (Not straight) Paths

Guys, So let's check out this problem here. We have hikers who start at the bottom of a mountain. So basically, I'm going to call this point A like this. And they're trying to get to the top of this mountain, so I'm going to call this point B. But instead of taking a straight path, like up like this, and therefore kind of looking like an inclined plane, what's actually happening is this hiker is going to take an irregular path with varying speeds and also inclinations. So what those would look like is kind of like a squiggly path to get up to the top here of this mountain. We want to do is we want to figure out the work that is done by gravity on the hiker during this entire hike. So basically, this is what I'm going to do. I'm going to draw a little line like this, and we know that this height here of the mountain, this is going to be my delta y, is going to be 1,000 meters. But the path that I'm taking from A to B is actually going to be sort of not straight. It's going to be curvy like this. So how do I calculate the work that's done by gravity? Well, remember that the equation is just

Wg = - mg × Δy

and what we have to do is we have to figure out a direction first of positive. So because I'm going from the bottom to the top, I'm going to call the upward direction positive like this. So what this means is that our delta y is actually positive 1,000 here. So if we get to the equation

Wg = negative mg × Δy

, we run into a problem here because you might think that as we go along this path here, the work that's done is going to be sort of changing, and you're actually right. It is. At any point along the path right here for this hiker, we know that the mg is going to point downwards. However, one of the really cool things about the work that's done by gravity is from the equation, we can tell that the work done only really depends on delta y. It's only dependent on the change in the height here. So the work that's done by gravity depends only on the change in your height, and it doesn't actually cause it doesn't actually depend on the path it's taken. So you wouldn't be able to solve this problem if you didn't know this. So this is actually called path independence. You might see that in your textbooks. It basically just means that it doesn't matter what path you take. All that matters is that you started at A and ended at B and the difference between those two is 1,000 meters. So the work that's done by gravity is just going to be negative. We have 75 times 9.8, and then we're going to multiply this by positive 1,000. So as we should expect, the work done by gravity is going to be negative, and it's going to be 735,000 joules. That's the work that is done by gravity. Alright? So again, only just depends on the height. It doesn't actually matter the path that you take here. We'll talk about that more, in a later video. So that's it for this one, guys.

Here’s what students ask on this topic:

What is the formula for calculating work in physics?

The formula for calculating work (W) in physics is given by:

$W=Fdcos\left(\theta \right)$

where:

• $F$ is the magnitude of the force applied.
• $d$ is the displacement of the object.
• $\theta$ is the angle between the force and the displacement vectors.

This equation shows that work is the product of the force, the displacement, and the cosine of the angle between them.

How do you calculate the work done by gravity?

To calculate the work done by gravity, use the formula:

$W{g}_{}=mg\Delta y$

where:

• $m$ is the mass of the object.
• $g$ is the acceleration due to gravity (9.8 m/s²).
• $\Delta y$ is the vertical displacement.

Gravity does positive work when an object falls and negative work when it rises.

What is the relationship between work and kinetic energy?

The relationship between work and kinetic energy is described by the Work-Energy Theorem, which states that the work done on an object is equal to the change in its kinetic energy:

$W=\Delta \mathrm{KE}$

where:

• $W$ is the work done.
• $\Delta \mathrm{KE}$ is the change in kinetic energy.

Kinetic energy (KE) is given by:

$\mathrm{KE}=\frac{1}{2}m{v}^2$

where $m$ is the mass and $v$ is the velocity.

How do you determine if work is positive or negative?

Work is positive when the force applied to an object aids its motion, meaning the force and displacement are in the same direction. Conversely, work is negative when the force opposes the object's motion, meaning the force and displacement are in opposite directions. Mathematically, this is determined by the cosine of the angle $\theta$ between the force and displacement vectors:

$W=Fdcos\left(\theta \right)$

If $\theta$ is between 0° and 90°, $cos\left(\theta \right)$ is positive, resulting in positive work. If $\theta$ is between 90° and 180°, $cos\left(\theta \right)$ is negative, resulting in negative work.

What is the significance of the angle in the work formula?

The angle $\theta$ in the work formula $W=Fdcos\left(\theta \right)$ represents the angle between the force vector and the displacement vector. It determines the component of the force that is in the direction of the displacement:

• If $\theta$ = 0°, the force is entirely in the direction of the displacement, and $cos\left(\theta \right)$ = 1, resulting in maximum positive work.
• If $\theta$ = 90°, the force is perpendicular to the displacement, and $cos\left(\theta \right)$ = 0, resulting in zero work.
• If $\theta$ = 180°, the force is opposite to the direction of the displacement, and $cos\left(\theta \right)$ = -1, resulting in maximum negative work.