Lesson Video: Work Done by a Constant Force | Nagwa Lesson Video: Work Done by a Constant Force | Nagwa

# Lesson Video: Work Done by a Constant Force Mathematics

In this video, we will learn how to calculate the work done by a constant force acting on a particle.

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### Video Transcript

In this video, we’ll learn how to calculate the work done by a constant force acting on a particle. As we’ll see, this work done can be positive, negative, or zero. This has to do with the direction a particle moves compared to the direction of the force acting on it.

To get started, let’s consider this situation of a box of some mass at rest on the ground next to a table. Say that we walk up to this box and then pick it up, applying a constant vertically upward force to do this. If we know the strength of that constant force, we’ll call it 𝐹, and the distance through which the box moved, we’ll call that 𝑑, then we can compute the work done on the box by our force 𝐹 according to this relationship.

In this equation, 𝑑 is technically a displacement rather than a distance. That’s important because it means the direction of 𝑑 is taken into account. In this case of us lifting up the box, the fact that 𝑑 and 𝐹 are in the same direction means that the overall work that we do on the box during this process is positive. We can represent that like this. When we apply a constant force 𝐹 to lift a box up, and that box does indeed go up, then the work done on the box by the force 𝐹 is positive. In terms of its magnitude, it would be the magnitude of the force 𝐹 multiplied by this distance 𝑑.

Next, let’s say that we place this box on top of the table. And let’s imagine further that this tabletop is smooth, that there’s essentially no friction between the box and the table. In that case, it would take literally no effort at all to push this box horizontally along the table. Here, the distance the box travels is to the right, while the force needed to create this motion is zero. This means that sliding our box across our smooth tabletop requires zero work.

But now let’s imagine a third motion for this box. Say we catch the box at the other side of the table. And then applying the same vertically upward force as before on the box, this time, we let it move downward at a constant speed under the force of gravity. Just like before then, the force we apply points up. But now the box moves down instead of up. And because the force and the box’s displacement are in opposite directions, this means the work that our applied force 𝐹 does on the box is actually negative.

As we’ve seen, the reason for this is that 𝐹 and 𝑑 are pointing in opposite directions. When it comes to work then, we do have this formula where we calculate work by multiplying 𝐹 and 𝑑. But we see that the outcome of this calculation depends on the relative directions of these two factors. And in fact, it’s only the parallel or antiparallel components of these factors that go into calculating work.

For example, say that we applied a force in this direction on our box. If the box moved only horizontally, as a result of this applied force, then we could say that it’s only the horizontal component of this applied force that did any work. The vertical component did no work. And we can see this because in this vertical dimension, the distance traveled 𝑑 is zero.

Now, it’s worth pointing out that even though we are talking about constant forces being applied to different objects, it’s still possible that this may accelerate the objects in question. For example, if we say that the floor, like our tabletop, is smooth so that this box slides frictionlessly across it, then that means there will be a nonzero net force in the horizontal direction. And this may remind us of Newton’s second law of motion, which says that the net force acting on an object in a given dimension equals that object’s mass times its acceleration in that dimension.

All this to say the force in our work equation may be an individual force or it may be a net force, or they may even be the same thing. This shows us that when we talk about work, it’s important to be specific about what force is doing that work.

Recall that when we were lowering our box from approximately here to here, the work we did by applying an upward force was negative because that force and the distance the box moved were in opposite directions. However, if we considered instead the work done by the force of gravity acting down on the box, in that case, we would have a downward-acting force with a downward-pointing distance. So the work done by gravity on the box would be positive.

Let’s get some practice with these ideas now through an example.

Calculate the work done by a force of 13 newtons acting on a body that moved 40 meters to the north if the force was acting towards the south. State your answer in joules.

Okay, so if we say that north points upward, like this, and we imagine we’re looking down on some body from above, then we’re told that this particular body moves 40 meters to the north and that all throughout that motion the body is subject to a 13-newton force acting southward. We want to calculate the work done by this 13-newton force.

We can recall that work is equal to force multiplied by displacement. Sometimes we hear this factor called a distance. But really, it’s important to keep in mind the direction of our object’s motion. A good example of that is our current situation. If we consider the northern direction to be positive, then we would say our body’s total displacement is positive 40 meters. This implies though that the southern direction is considered negative. And so we would say that the 13-newton force, we’ll call it 𝐹, is negative 13 newtons.

All this means that when we use this value of 𝐹 and 𝑑 in our equation for work, we have negative 13 newtons being multiplied by positive 40 meters. A newton times a meter is a joule. And so our answer comes out to negative 520 joules. This is the work done by the 13-newton force acting on this body.

Now let’s look at an example involving Newton’s second law of motion.

A force acted on a body of mass 400 grams which had been at rest, causing it to accelerate at 36 centimeters per second squared. If the work done by this force was 0.72 joules, find the distance the body moved.

Alright, so let’s say that this is our body, which starts out at rest, and then a force is applied to it. So it starts to accelerate. And we’ll call this acceleration 𝑎. We’re told the mass of our body, we’ll call that 𝑚, as well as the work done on it, we’ll call it 𝑊, by this force. Knowing that this work was done while the body moved across some distance, we’ll call it 𝑑, it’s that distance we want to solve for.

To start doing this, we can recall that work is equal to force multiplied by distance. Strictly speaking, this 𝑑 is a displacement. But in this case, we can treat it as a distance, a quantity with no direction associated with it. Along with this equation for work, we can recall Newton’s second law of motion. It tells us that the net force acting on an object is equal to that object’s mass multiplied by its acceleration.

In this exercise, we’re told that this force that acts on our body is what causes it to accelerate at this given rate. This means the force does represent the net force acting on the body. And so we can replace 𝐹 in our equation for work with 𝑚 times 𝑎 from Newton’s second law. In this equation, it’s the distance 𝑑 that we want to solve for.

If we divide both sides of the equation by 𝑚 times 𝑎, those factors cancel on the right. And we find that 𝑑 equals 𝑊 over 𝑚 times 𝑎. If we substitute in the given values of 𝑊, 𝑚, and 𝑎, we find that, because of the units involved, we’re not quite ready to calculate 𝑑. We’ll want to convert the mass of our body to the SI base unit of kilograms and our body’s acceleration to units of meters per second squared.

Recalling that 1,000 grams equals one kilogram and that 100 centimeters is one meter, we can write an equivalent expression in updated units like this. Our body’s mass is 0.400 kilograms, while its acceleration is 0.36 meters per second squared. At this point, all of our units agree. So we can calculate 𝑑. We find it to equal exactly five meters. This is the distance our body moved while accelerating under the given conditions.

Let’s look now at another example.

A construction worker of mass 100 kilograms is carrying some bricks up a ladder of height 15 meters. If the work done by the construction worker in climbing the ladder is 20,433 joules, find the mass of the bricks. Take 𝑔 to equal 9.8 meters per second squared.

Okay, so let’s say that this is our ladder, which we’re told has a height we’ll call ℎ of 15 meters, and that on this ladder there’s a construction worker with a mass we’ll call 𝑚 sub 𝑐, who has climbed all the way to the top carrying a load of bricks with a mass we’ll call 𝑚 sub 𝑏. Knowing that this construction worker did 20,433 joules of work in climbing this ladder with this load of bricks, we want to determine the mass of the bricks, 𝑚 sub 𝑏.

If we think about the combined mass involved, the mass of the bricks plus the mass of the construction worker, we know that this total mass multiplied by the acceleration due to gravity is equal to the weight force that acts on this worker. But then, over against this force, the construction worker is able to climb up a height of 15 meters. We can say then that the magnitude of the force the construction worker exerts, we’ll call it 𝐹 sub 𝑐, is equal to this total mass times 𝑔.

At this point, we can recall that the work done by a given force 𝐹 is equal to that force multiplied by the displacement of an object on which the force acts. In our case, our body’s displacement is the height of our ladder ℎ, while the force involved is the sum of the masses of the bricks and the construction worker times 𝑔. Therefore, we can write this expression for the work done by the construction worker. Recalling that it’s 𝑚 sub 𝑏 that we want to solve for, we divide both sides of the equation by 𝑔 times ℎ. Those factors then are canceled on the right. And following that, we subtract 𝑚 sub 𝑐, the mass of the construction worker, from both sides. That results in this equation for the mass of the bricks. When we plug in for the values of 𝑊, 𝑔, ℎ, and 𝑚 sub 𝑐, we obtain a result of 39 kilograms. This is the mass of the bricks the worker carried to the top of the ladder.

Let’s look now at one last example involving work.

A man of mass 94 kilograms ascended a plane of length 90 meters, which was inclined at an angle of 30 degrees to the horizontal. Determine the work done by his weight to the nearest joule. Take 𝑔 to equal 9.8 meters per second squared.

Okay, so let’s say that this is our plane at 30 degrees to the horizontal. And there’s a man of mass 94 kilograms climbing up the plane. We’ll call that mass 𝑚. And we’re told that the length of the plane is 90 meters, which we’ll call 𝑙. We want to calculate the work done by this man’s weight to the nearest joule.

We can say then that this man’s weight, his mass times the acceleration due to gravity, a downward-acting force, actually does work as this man climbs up the incline. That work is equal to the force involved, that is, the man’s weight force, multiplied by the vertical distance through which the man moves. That is, the distance we’ll use in our calculation won’t be the length 𝑙. But rather it will be this vertical distance 𝑑. That’s because this distance represents the only component of the length 𝑙, we could call it, that is parallel or antiparallel to the force involved.

Here’s where we can start then. We know that the force for which we want to calculate work done is the man’s weight force, 𝑚 times 𝑔. Going a step further, since this 30-degree slope is part of a right triangle, we can say that the distance 𝑑 is equal to 𝑙 times the sin of 30 degrees. This is so because the sine of this angle is equal to the ratio of 𝑑 to 𝑙. Rearranging that relationship, we get that 𝑑 equals 𝑙 sin 30.

Before we plug in the given values for 𝑚, 𝑔, and 𝑙, it’s important to set up a sign convention. Note that as this man climbs the incline, he moves upward vertically. This is in the opposite direction to his weight force, which acts down. To make sure that we combine quantities correctly, let’s say that one of these directions is positive, making the other negative. If we say that the upward direction is positive, then that means the value of 𝑙 times the sin of 30 degrees is positive. But then that means that 𝑚 times 𝑔, which is directed downward, must be negative. The point here is that the force involved, the man’s weight force, is acting in an opposite direction to the displacement 𝑑. The net effect of this is that the work done by this weight force will be negative.

Now that we’ve figured out the signs, we’re ready to substitute in for 𝑚, 𝑔, and 𝑙. The man’s mass is 94 kilograms, 𝑔 is 9.8 meters per second squared, and 𝑙 is 90 meters. Entering this expression on our calculator, we get a result, to the nearest joule, of negative 41,454 joules. This, to the nearest joule, is the work done by this man’s weight as he ascends this plane.

Let’s finish up now by summarizing a few key points. In this video, we’ve seen that the work done by a constant force 𝐹 on some object is equal to the product of that force and the displacement that the object undergoes. For both 𝐹 and 𝑑, the direction is very important. In general, the work done by a constant force can be positive, negative, or zero. Lastly, we saw that it’s only the parallel or antiparallel components of 𝐹 and 𝑑 that contribute to the total work done by the force 𝐹.