A skydiver falls at a constant speed with their parachute not yet opened. For all parts of this question, use 9.8 meters per second squared for the acceleration due to gravity. Which of the following statements is a description of Newton’s third law of motion? Tick one box. The acceleration of an object is proportional to its mass and to the force applied to it. When a force is applied to an object, the object exerts an equal force in the opposite direction to the applied force. An object only changes its velocity due to forces applied to it.
To figure out which one of these three statements gives the best description of Newton’s third law of motion, let’s recall those three laws. The first law of motion has to do with an object’s inertia, its tendency to keep moving or keep still if that’s what it’s doing already.
Of our three statements, the last one — it says an object only changes its velocity due to forces applied to it — is the best match for this law. So we know it’s not a good match for the third law of motion.
Let’s move on to the second law. This law of motion is often summed up as an equation that the net force on an object is equal to its mass times its acceleration. Considering our first two answer options, it’s the very first one which uses terms that show up in the second law: object acceleration, mass, and force applied to the object. This first statement then suggests the second law of motion, but not the third.
So then on to the third and final law of motion. We can summarize this one by saying that every action has an equal and opposite reaction to it. Let’s look again at the second statement. It says when a force is applied to an object, the object exerts an equal force in the opposite direction to the applied force. That sounds a lot like the third law, so we’ll tick that box.
As we go on in this exercise, we’ll make use not only of this law of motion, but of the other two as well. Let’s keep going. Figure one shows the direction and magnitude of the skydiver’s weight. Show the other force acting on the skydiver.
Looking at figure one, we see that indeed this is true. There’s a weight force vector drawn in on this diagram, with the direction pointing down and the magnitude of 588 newtons. Now looking at this diagram, we might think that the only force acting on the skydiver is their weight force.
But there’s a vital piece of information that was given about this scenario earlier. We were told at the outset that as the skydiver descends, before their parachute is deployed, as we see in this figure, they’re moving with a constant velocity.
Now if an object is moving with constant velocity, what does that mean? Well, it means that it has no acceleration or that the acceleration of the object is zero. And if the acceleration of an object is zero, that means something about the net force acting on it.
Remember the second law of motion that we saw earlier. This is a direct application of that law. If the acceleration, 𝑎, is zero, then that must mean the left side of the equation is zero too. In other words, the net force is zero. And if the net force is zero, what does that mean about figure one? It means that it’s incomplete. There must be some other force acting on the skydiver to cancel out the weight force we see drawn in. In order to effectively cancel out the weight force, it must be in the opposite direction and must be equal in magnitude.
Now we may wonder, just what force is this that’s acting upward with the same magnitude as the weight force? Well, it’s the force of air resistance that acts on the skydiver as they’re falling. And at this point in time, that air resistance force perfectly balances the weight force of the skydiver.
We can move on now to the next part of the exercise. But before we do, let’s start a bank of information that we’ll keep on the left side of our screen. This will be a way of keeping track of all the information we collect as we go along. For example, we were told that the acceleration due to gravity is to be treated as 9.8 meters per second squared. We’ll put that information here. And we also know now that the weight force acting on the skydiver is 588 newtons. So we’ll record that as well. We’ll keep adding to our information bank as we go, but we’re now ready to move on to the next question.
Calculate the mass of the skydiver.
Okay, to solve for this mass, let’s take stock of what we have in terms of related information. We know both the weight of the skydiver and we know the acceleration due to gravity that the skydiver experiences. What’s more, we can recall a mathematical relationship that connects these three values together.
Weight, being a force, is equal to the mass of an object multiplied by the acceleration it experiences due to gravity. Now in our case of course, we want to solve not for weight but for mass. So to isolate mass on one side of the equation, we’ll divide both sides by the acceleration due to gravity, 𝑔. When we do that, on the right-hand side of our equation, that factor cancels out. And we see that the mass of our skydiver is equal to their weight divided by 𝑔.
In our information bank, we have both these bits of information, so we’ll plug them in to solve for 𝑚. 𝑚 is equal to 588 newtons divided by 9.8 meters per second squared, or 60 kilograms. That’s how the units of this expression simplify. So that’s the mass of our skydiver, and we’ll add this information to our growing collection.
Next, let’s look into how the skydiver accelerates. The skydiver deploys their parachute, which increases the air resistance. After the skydiver’s parachute is deployed, their velocity decreases. In 3.5 seconds, their velocity changes from 30 meters a second to 5.5 meters a second. Calculate the acceleration of the skydiver.
So we’re talking here about acceleration, which we can recall is related to velocity change and time change. Specifically, the acceleration of an object is equal to its final velocity minus its initial velocity, all divided by the time interval over which this velocity changes.
Now in our case, that final velocity is 5.5 meters a second and the initial velocity is 30 meters a second. And this all happens over a time span of 3.5 seconds. What was it that caused this change in velocity? Initially, just as our skydiver was deploying their parachute, they were falling downward with their initial velocity 𝑣 sub 𝑖. But then as they deploy their parachute, over three and a half seconds, they slow down to a final velocity we’ve called 𝑣 sub 𝑓. That’s the reason for their velocity change; their parachute has opened up.
When we calculate their acceleration over this time, we get a negative value, negative seven meters per second squared. This makes sense if we think about it because the acceleration due to gravity, 𝑔, is a positive value and we know that points down. Therefore, an acceleration upward, which our skydiver has when their parachute deploys, will be considered negative, and that’s what we found. Over this time interval, the skydiver’s acceleration is negative seven meters per second squared. And we’ll add that information to our growing list.
Next, let’s look at the force that this open parachute subjects the skydiver to. Calculate the upward force due to the increased air resistance that the parachute produces.
So here’s the idea. While the skydiver was falling, there was an air resistance force pushing up on the skydiver. Then when the parachute is opened up, there’s a much larger air resistance force acting on the skydiver and parachute combined. This is what makes the skydiver slow down and is able to approach earth at a safe speed.
Now we want to calculate the force that’s due only to the increased air resistance as the parachute is opened. In other words, we won’t take into account the air resistance experienced by the skydiver before they deploy their parachute.
Looking over at our longer air resistance vector, we can almost imagine that a portion of it is due solely to the parachute having opened up. And then it’s that portion we want to calculate. And to do it, we can be helped by Newton’s second law of motion. This law tells us that the net force on an object is equal to the mass of that object times its acceleration. And in our case, this increase in force, which we can call 𝐹 sub 𝑅, for the force due to air resistance, is equal to the skydiver’s mass multiplied by their acceleration after the chute has opened.
Now we have values for both the mass and the acceleration of the skydiver we solved for previously. So we can plug those in now. Notice, when we do that, if we multiply these numbers together as is, we’ll get a negative force. Now in some sense, this is an accurate result because our resistive force is pointing upward, which by our convention is in the negative direction. Yet if we just hold the understanding that this is an upward force and therefore implicitly negative, we can report our force as a positive overall value. We’ll essentially report force magnitude.
Multiplying the mass times the acceleration, we find a result of negative 420 newtons. But when we report this as a magnitude, that is, a value without regard to its sign, then we’ll simply say this upward force is 420 newtons. That’s the force magnitude on the skydiver due to the increased air resistance they experience. Adding this also to our list of information, let’s now move on to consider the distance traveled by the skydiver as they’re falling.
Figure two shows the change of the skydiver’s downward velocity with time from when they open their parachute. Use the graph to determine the distance that the skydiver falls in the 3.5 seconds after their parachute opens.
Seeing figure two, it shows us the velocity of the skydiver compared to time starting at the time when the parachute of the skydiver opens. So for this entire 3.5-second time interval, the skydiver’s parachute is open and their velocity is going down, down, down. What we want to do is figure out just how much distance the skydiver covers over this time interval.
Now because distance is equal to velocity times time, we’ll solve for that distance by multiplying the velocity on this curve by the time value of that velocity. This involves solving for the area under this velocity–time curve.
We can break this area up into two separate chunks: one the area of a triangle we’ll call 𝐴 sub triangle and the other the area of a rectangle we’ll call 𝐴 sub rectangle. The distance that the skydiver falls we want to solve for is equal to the sum of these two areas. That’s equal to one-half the base of the triangle times its height plus the base of the rectangle times its height.
For both shapes, the base is 3.5 seconds. And for the rectangle, the height is 5.5 meters per second. And for the triangle, the height is 30 minus 5.5 meters per second. Calculating all this out, we find a result of 62.125 meters. That’s how far the skydiver fell during these 3.5 seconds. Let’s move on now to consider the work involved here.
Calculate the work done on the skydiver by the parachute.
To solve for this work, let’s review what’s going on. Once the skydiver has deployed their parachute, there’s an increased resistive force due to air resistance we’ve called 𝐹 sub 𝑅. That force is a constant force as the skydiver descends a distance 𝐷 with their parachute deployed.
Now if we recall the fact that work is equal to force multiplied by distance, we see that we have all the ingredients needed to calculate this work done. The work done on the skydiver by the parachute is equal to 𝐹 sub 𝑅, which we solved for in an earlier part of the exercise, multiplied by 𝐷, which is what we just found. The work done equals 420 newtons times 62.125 meters, or 26092.5 joules. That’s the work done on the skydiver by the parachute.
Next, let’s consider the kinetic energy involved in this process. Calculate the skydiver’s kinetic energy before and after their deceleration.
Now we can recall an equation that connects kinetic energy with the skydiver’s mass and velocity. This relationship tells us that an object’s kinetic energy is equal to one-half its mass times its velocity squared. In the case of our skydiver, we know their mass — that’s given as 60 kilograms — but what is their velocity before and after the deceleration?
To recall those values, let’s bring back up figure two. Here, once more, is figure two. And critically, it shows us the skydiver’s velocity right before and after deploying their chute. That is, it shows us the velocity of the skydiver before and after their deceleration. We can use these values then to calculate the kinetic energy of the skydiver both before and after this deceleration.
So then KE sub 𝐵, what we call the kinetic energy of the skydiver before they decelerate, is equal to one-half their mass of 60 kilograms multiplied by 30 meters per second squared. This works out to the nice round number of 27000 joules.
Then what’s the skydiver’s kinetic energy after deceleration, that is, when their speed is 5.5 meters per second? We’ll call that KE sub 𝐴. And it’s equal to one-half 60 kilograms multiplied by 5.5 meters per second squared. This results in the much smaller number of 907.5 joules. So we see the kinetic energy of the skydiver has decreased significantly over the course of this deceleration.
Finally, let’s tie this all together by comparing these kinetic energies to the work done on the skydiver. Explain the relationship between the work done on the skydiver by the parachute and the skydiver’s kinetic energy before and after their deceleration.
Okay, to start off, let’s recall what those values were. The work done we calculated as 26092.5 joules. And just previously, we solved for the kinetic energy of the skydiver before and after their deceleration, with those values right here.
Now here’s something interesting. Let’s say we solve for the difference between these two kinetic energies. That is, we subtract KE sub 𝐴 from KE sub 𝐵. Now if we do that and calculate this difference, we find a result of 26092.5 joules. No big deal, right?
Well, wait a second. This value is exactly the same as the value we got for the work done on the skydiver by the parachute. Is that just a coincidence? Oh, it turns out it’s not. Here’s what we’re seeing. We’re seeing that the difference in kinetic energy of the skydiver before and after deceleration is equal to the work done on them.
Now we can write this difference in kinetic energy as ΔKE. So what we’re saying is ΔKE in our case is equal to the work done. And in fact, this is generally true of physical processes. We have sort of, kind of accidentally discovered a theorem known as the work energy theorem, which is really just a fancy expression for the equation we have here, which says that the change in an object’s kinetic energy is equal to the work done on it.
Speaking of work, we recall that work is itself a form of energy. And a clue to that is that the units of work are the same as the units of energy, joules. So then this statement that we have here boxed in green is really a statement of energy conservation.
We’re essentially saying that the loss in kinetic energy we experienced is equal to the work done on the skydiver. So then this is the relationship between the change in kinetic energy of the skydiver and the work done on them. And as a corollary, we can say that this is equivalently saying that energy is conserved. This is how these two quantities, the change in kinetic energy of the skydiver and the work done on them, are related.