Question Video: Acceleration and Force | Nagwa Question Video: Acceleration and Force | Nagwa

# Question Video: Acceleration and Force Physics • First Year of Secondary School

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A swimmer of mass 45 kg uses her legs to push herself away from a swimming pool wall, applying a force of 280 N. The water that the swimmer accelerates through applies 160 N of force in the opposite direction to that in which she accelerates. What is the swimmer’s acceleration through the water?

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### Video Transcript

A swimmer of mass 45 kilograms uses her legs to push herself away from a swimming pool wall, applying a force of 280 newtons. The water that the swimmer accelerates through applies 160 newtons of force in the opposite direction to that in which she accelerates. What is the swimmer’s acceleration through the water?

Okay, so in this question, we’ve got a swimmer moving through some water. So here’s our swimmer. And we’ve been told that she uses her legs to push herself away from a swimming pool wall. So let’s say that this is the wall that she’s pushing off from. And just for fun, we can draw in the surface of the water.

Anyway, so we’ve been told that the swimmer firstly has a mass of 45 kilograms. Let’s label her mass 𝑚 and say it’s 45 kilograms. We’ve also been told that the swimmer pushes herself off the wall, applying a force of 280 newtons. In other words, the swimmer is exerting a 280-newton force on the wall. But what this means is that we can recall Newton’s third law of motion.

What Newton’s third law of motion tells us is that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A. Now in our situation here, the swimmer is object A. It’s exerting a 280-newton force on object B, which is the wall. Therefore, Newton’s third law of motion tells us that the wall, object B, will exert an equal, 280 newtons, and opposite, to the right as we’ve drawn it, force on the swimmer, 280 newtons.

Now this is important because, as well as this 280-newton force on the swimmer to the right as we’ve drawn it, we’ve been told that the water exerts a 160-newton force on the swimmer in the opposite direction to that in which she accelerates. Well, the swimmer is accelerating to the right because she’s trying to push away from the wall. And of course, once again, we’ve just arbitrarily drawn the wall to the left of the swimmer. We could’ve equally drawn the swimmer going in the opposite direction and the wall to the right. But the point is that the swimmer is trying to accelerate away from the wall.

And so we can say that her acceleration is to the right as we’ve drawn it. And we’ll call this acceleration 𝑎. But then the water we’ve been told exerts a 160-newton force on the swimmer in the opposite direction to which she’s accelerating. This means that there are now two forces acting on the swimmer: the 280-newton force to the right because of the wall and 160-newton force to the left because of the water.

We can simplify all of this by drawing the swimmer just as a blob and considering that there’s a 280-newton force to the right and a 160-newton force to the left. Then we can account for these two forces and find the resultant or overall force on the swimmer. This overall force is also called the net force. And if we consider arbitrarily that the direction to the right is positive and the direction to the left is negative, then we can add these two forces together.

And the net force, which we’ll call 𝐹 subscript net, is equal to the positive 280 newtons, cause that force is to the right, minus the 160 newtons, which is to the left. And that net force, therefore, ends up being 120 newtons. Therefore, it’s a 120-newton force to the right. And this kind of makes sense. If there’s a 280-newton force to the right and a 160-newton force to the left, then 160 of those 280 newtons to the right are being cancelled by this force to the left. And whatever remains — that’s the 120 newtons — is going to be the resultant force to the right.

If, however, we had arbitrarily chosen the left to be positive and the right to be negative, then we’d have found the net Force, 𝐹 subscript net, to be the positive 160 newtons minus the 280 newtons. And that would’ve given us a net force of negative 120 newtons. So what gives?

Well, the reason for this is because we chose this way, to the left, to be positive. And hence, what we found is that the net force is negative 120 newtons to the left. Or in other words, it’s 120 newtons to the right, exactly as what we found here. So even though we do need to be careful with signs, regardless of which convention we chose, we’d find the net force to be 120 newtons to the right.

So now that we found the net force on our swimmer, we can recall 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 multiplied by the acceleration it experiences. And since we’ve already found the net force and we’ve been given the mass of the swimmer, we can find out the acceleration experienced by the swimmer.

We can do this by rearranging the equation by dividing both sides of the equation by the mass, which results in the mass cancelling out on the right. This way, what we’re left with is that the net force on the swimmer divided by the mass of the swimmer is equal to the acceleration experienced by the swimmer. So we can say that the acceleration, 𝑎, is equal to the 120-newton net force to the right divided by the mass of the swimmer, which is 45 kilograms.

We can also see that we’re working in base units because the base unit of force is the newton and the base unit of mass is the kilogram. And so our answer for the acceleration is going to be in its own base unit, which is metres per second squared. And so when we evaluate this fraction on the right-hand side, we find that the acceleration is equal to 2.6 recurring. And the unit is metres per second squared.

Now because the smallest number of significant figures that we’ve been given in the question is two significant figures, we can give our final answer to two significant figures as well. Now 2.6 recurring is basically two point and then six occurring many times. And the reason we write it like this is so that we can work out that this is the first significant figure and this is the second.

Well, the second significant figure is either going to stay the same or it’s going to round up, depending on what the next significant figure is. The next one is a six. Since six is larger than five, this second significant figure is going to round up. And so our value becomes 2.7 metres per second squared. So at this point, we have our final answer. The swimmer’s acceleration is 2.7 metres per second squared.

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