Video: Mechanical Work

In this lesson, we will learn how to calculate the work done by a force that acts on an object over a distance.

11:52

Video Transcript

In this video, we’re talking about mechanical work. In our everyday life, when we use this word “work,” we might be talking about any number of things. When we’re doing any homework or say chores about the house, we say we’re doing work. And then when mom or dad goes off for the day and comes back, we say they were at work. In the world of physics, though, this term has a very specific meaning.

Here’s the definition for work that we’ll use. It’s the product of the force applied to an object and that object’s displacement. In other words, we find the work done on an object by multiplying the force applied to it with that object’s straight-line distance traveled. Based on this definition, we can write an equation for work, where work is typically symbolized using a capital 𝑊. We’ve said that work is the product of force and displacement, which means we can write that work is equal to force 𝐹 multiplied by displacement 𝑑. Here’s an example of work going on.

Let’s say we have a car traveling down the road. The car is moving left to right. What’s happening here is the road surface is pushing on the tires of the car in such a direction that it moves the car forward. We could say that the total force the road applies to these car tires is 𝐹. And if — thanks to this force 𝐹 — the car moves forward a distance we can call 𝑑, then we can say that the work done on this car — we’ll call it 𝑊 sub 𝑐 — is equal to that force applied by the road onto the car multiplied by the distance the car travels.

Imagine we had a box. We can say this box has a mass 𝑚 and it’s stationary on a table top. Then let’s say we pick up this box. And we lift up this box some distance — we can call this 𝑑 — above this table level. What we’ve done is we’ve applied an upward force on the box. That’s the force here in our equation for work. And we’ve done it over a certain distance 𝑑. In lifting this box then, we’ve done work on it. So those were some examples of processes where work takes place. But to better understand this equation, let’s look at the terms involved in it one by one.

We can start with the force involved. We know that the SI base unit of force is the newton, symbolized capital 𝑁. And we can recall that one newton is the force that’s needed in order to accelerate a one-kilogram mass with an acceleration of one meter per second squared. In other words, the unit of newton is equal to a kilogram meter per second squared. And then next, we can consider this displacement 𝑑, also known as the straight-line distance traveled by our object. The SI base unit of displacement or distance is meters. Based on the units involved then for force as well as displacement, we can see that when we multiply a force by a displacement, we’ll have units of newtons multiplied by meters. These units show us that indeed when we talk about work, we’re talking about forces over distances.

Now, if we multiply a single newton of force by a single meter of displacement, then that product between these two is its own unit. It’s called the joule. This unit, which is abbreviated simply as a capital 𝐽, is the unit for work. So as we’ve seen, the unit of force is newton and the unit of displacement is meters. And if we multiply a newton by a meter, we get this unit called a joule and that’s the unit of work. We could say then that some process might involve 100 joules of work or that a person might do 250 joules of work lifting a box. This is the way we talk about work done.

So now, given a scenario where we have a force in newtons and a displacement in meters, we’re able to calculate work done in units of joules. But there are a couple of things to watch out for when applying this particular equation. For one thing, the equation assumes that the force we’re talking about is a constant force. It doesn’t change over time. That means it always has the same magnitude or strength, and it also always points in the same direction. Another assumption we’re making here is that the force and the displacement involved are in the same direction. For example, if we have some object with a force acting on it and say that force is acting to the right, then this equation — written as it is — assumes that the object’s displacement is also in the same direction as the force acting on it.

So those are two things to watch out for as we apply this equation for work done: that the force involved is constant and that the force and the object displacement point in the same direction. Now the very best way to get some expertise with this relationship is to put it into practice. Let’s do that now through a couple of example exercises.

Which of the following formulas correctly shows the relationship between the magnitude of a force, the work done by that force, and the distance moved in the direction of the applied force by the object that the force is applied to?

Before we review our answer options, let’s consider what’s going on in this situation. We have some object. We’ll say that this is our object. And we imagine that some force is being applied to this object. We can call that force 𝐹. And that along with this, the object moves or is displaced at some distance — we can call that 𝑑 — in the same direction as this force 𝐹 is applied. Based on that, our question is, which of these five options — A, B, C, D, or E — gives us the correct formula, showing the relationship between these two quantities — force, distance — and a third quantity the work done on our object. In other words, we want to know how work, force, and distance are related to one another mathematically.

That being said, let’s look at our answer candidates. A) Work is equal to distance divided by force. B) Work is equal to force divided by distance. C) work is equal to distance minus force. D) Work is equal to force times distance. E) Work is equal to force minus distance.

One of the first things we can do to start narrowing down our answer options is to consider the units involved in these expressions. Each one of these expressions involves three terms. There’s work, 𝑊, force, 𝐹, and distance, 𝑑. Now, if we consider the units of each of these three terms, work, force, and distance, then starting at the top, the unit of work is the joule, abbreviated capital 𝐽. The base unit for force we know to be the newton, abbreviated capital 𝑁. And the base unit for distance or displacement is the meter.

Now, let’s consider for a moment that the unit of work is the joule. Looking at these five answer options, we see that work appears on the left-hand side of each one. And since the unit of work is the joule, that means the left-hand side of every one of these equations has units of joules. Now, if some amount of work in units of joules is equal to some other quantity, that is the quantity on the right-hand side, then that means the right-hand side of the correct formula must also be expressed ultimately in these same units, in units of joules. This means that if we can say for sure that the units on the right-hand side of any of these expressions cannot be joules, in that case, those answer options cannot be the correct formula.

Keeping this in mind, let’s consider answer options C and E. Option C claims that work is equal to distance minus force, where option E claims that work is equal to force minus distance. We know that the units of force are newtons, and the units of distance are meters. But that means that we’re not able to combine the units of these terms in this way and wind up with units of joules, which we would have to, if they were to agree with the left-hand side. For example, for answer option C, we would have a distance in meters minus a force in newtons. But subtracting some number of newtons from some number of meters doesn’t give us some number of joules. The units don’t work out. And this means that answer option C can’t be our correct formula.

Answer option E drops out for the same reason. Some number of newtons minus some number of meters cannot equal some number of joules. So we’re down to answer options A, B, and D. Now, at this point, we can recall that there is a relationship that ties together the units of joules, newtons, and meters. The definition of the unit joule is that one joule is equal to one newton of force multiplied by one meter of distance. Written another way, we can say that a joule is equal to a newton meter. This shows us that whatever our correct answer option is, we’ll have units of newtons multiplied by meters on the right-hand side of our expression.

Looking at option A, we see that this candidate has units of meters divided by newtons, so not newtons times meters, whereas option B has units on the right-hand side of newtons divided by meters. Then, finally, option 𝑑 has units on the right-hand side of newtons multiplied by meters. We see that it’s this option which gives us the units on the right-hand side which agree with the units equivalent to a joule. Therefore, option D has the correct relationship between the units on the left-hand and right-hand side. And therefore, this is our choice for the formula correctly representing the relationship between work, force, and distance. Work is equal to force multiplied by distance. Now let’s look at a second example involving mechanical work.

A force of 320 newtons is continually applied to push a trolley through a supermarket parking lot. If the trolley is pushed for a distance of 15 meters, how much work was done on the trolley?

Now the first thing we can say here is that trolley in this case is another word for shopping cart. So we have this cart and we’re pushing it around the parking lot. We’re told specifically that as we push the cart, we’re applying a force — we can call it 𝐹 — of 320 newtons. And that thanks to that force being applied to the cart, the cart travels a distance — we can call that distance 𝑑 — given as 15 meters. What we want to find out is given this force and given this distance, how much work was done on the cart? To figure this out, let’s recall the relationship that connects work with force and distance.

Under two conditions and we’ll get to what those are in just a second. The work done on an object, 𝑊, is equal to the force applied to the object multiplied by the distance that it moves. Now here are those two conditions we mentioned. In order for this equation to be valid, the force 𝐹 must be a constant force. It can’t vary getting smaller or larger. And the second condition is that the force 𝐹 and the distance 𝑑 that the object travels must be in the same direction. As we look at our scenario, we see that both these conditions are met.

We have a constant force of 320 newtons and the force points in the same direction as our cart moves. This means we can indeed solve for the work done on this cart by multiplying the force applied to it with the distance that it travels. That work done on the cart — we can call it 𝑊 sub 𝑐 — is equal to the force on the cart, 320 newtons, multiplied by the distance it travels, 15 meters. Since both our force and our distance are already in their SI base units, newtons and meters, respectively, we can simply multiply these two values together with their units to solve for the work in units of joules. When we do, we find that 𝑊 sub 𝑐 is 4800 joules. That’s how much work was performed on this cart. Let’s take a moment now to summarize what we’ve learned in this lesson on mechanical work.

First off, we saw that work is equal to the force exerted on an object multiplied by the distance that object travels. Written as an equation, we can say that work 𝑊 is equal to force 𝐹 multiplied by distance 𝑑. We learned about the units involved in this expression that the base units of force is newtons, the base unit of distance is meters, and the base unit of work is something called joules, abbreviated capital 𝐽. And we learned that a joule is equal to a newton multiplied by a meter. Or, put another way, a joule of work is equal to a newton of force exerted over a meter of distance.

And lastly, we saw that this equation we learned for work — that work is equal to force times distance — has two requirements in order to be valid. First, there’s a requirement that the force involved must be constant. And the second condition is that the force involved and the displacement or distance traveled by the object need to point in the same direction. When those conditions are met, we can use this mathematical relationship to calculate the work done on an object.

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