Question Video: Motion-Resistive Forces | Nagwa Question Video: Motion-Resistive Forces | Nagwa

# Question Video: Motion-Resistive Forces Physics

A bat of mass 250 g flies horizontally, beating its wings to push the air back. The force applied by the bat’s wings is 2.5 N, and the bat accelerates at 6.2 m/s². How much drag force acts on the bat?

05:11

### Video Transcript

A bat of mass 250 grams flies horizontally, beating its wings to push the air back. The force applied by the bat’s wings is 2.5 newtons, and the bat accelerates at 6.2 meters per second squared. How much drag force acts on the bat?

Okay, so this question, we know that we’ve got a little bat with a mass of 250 grams, which is flying horizontally. So let’s say that this is our bat and it’s flying in this direction. Now we’ve been told that the bat is beating its wings to push the air back. This means that, due to the beating of the bat’s wings, the force on the air is to the left, because it’s pushing the air back.

And so by Newton’s third law of motion, every action has an equal and opposite reaction. And so the air is exerting an equal and opposite force on the bat. In other words, because the bat’s wings push the air back, the air pushes the bat forward.

Now this forward force we’ve been told, the force applied by the bat’s wings, is 2.5 newtons. So let’s label the force 2.5 newtons. And we’ve also been told that the bat accelerates at 6.2 meters per second squared. So let’s say that the acceleration of the bat is 6.2 meters per second squared. Finally, we’ve been asked to calculate the drag force acting on the bat.

So we know that there must be a force in the opposite direction because, of course, the drag force acts to oppose motion. In other words, if the bat is traveling this way, then the drag force is going to act in this direction. And we’re trying to find out the value of the drag force. So let’s call it 𝐹 subscript 𝑑 for force subscript drag. So how are we going to go about doing this?

Well, the first thing that we can do is to recall Newton’s second law of motion. This law tells us that the net force on an object, 𝐹 sub net, is equal to the mass of that object multiplied by the acceleration experienced by that object. Now in this case, our object is the bat, and we know the mass of the bat and the acceleration of the bat. The mass is 250 grams, and the acceleration is 6.2 meters per second squared. So we can work out the net or overall force on the bat.

To do this, we simply substitute in the values of the mass and the acceleration. But before we do, we need to notice that the value of the bat’s mass has been given to us in grams, whereas we need it in the standard unit of mass, which is kilograms. So to convert from grams to kilograms, we need to recall that one gram is equivalent to one thousandth of a kilogram. And then multiplying both sides of the equation by 250, we can see that 250 grams is equivalent to 0.25 kilograms.

Now this means we have our mass in kilograms and so we can substitute it in. So we say that the net force on the bat is equal to the mass of the bat in kilograms multiplied by the acceleration of the bat in meters per second squared, again the standard unit. Now because we’ve used standard units for these quantities, this means that our answer for the net force is also going to be in its standard unit. And of course, the standard unit for force is the newton.

And when we evaluate the right-hand side of the equation, we find that the net force on the bat is 1.55 newtons. This is the overall force on the bat which results in an acceleration of 6.2 meters per second squared for a bat which has a mass of 250 grams.

Now as we’ve already said, we’ve just calculated the net or the overall or resultant force on the bat. But how does this relate to the force of 2.5 newtons exerted by the bat on the air and also on the drag force? Well, we know that these two forces add together, of course taking into account the directions in which they’re acting, in order to give us a resultant of 1.55 newtons. And of course, that 1.55-newton force is acting in this direction, because the net force always acts in the same direction as the acceleration.

So to make life easy for ourselves, let’s say that this direction, to the right, the direction at which the bat is flying, is positive and the other direction is negative. This means that we can say- this means that we can add together the forces acting on the bat to give us the resultant force whilst accounting for their directions.

So first of all, the drag force, it’s acting towards the left of course, against the direction of motion. So we can say that negative 𝐹 sub 𝑑, because it’s acting to the left, plus the 2.5-newton Force, because it’s acting to the right, is equal to the resultant force, 1.55 newtons. And we get this equation because there are only two forces acting on the bat, the drag force and the force exerted by the air on the bat to move it forward, which means that all we need to do is to rearrange the equation now.

If we add 𝐹 sub 𝑑 and subtract 1.55 newtons from both sides of the equation, then on the left-hand side, the 𝐹 sub 𝑑s cancel, and on the right-hand side the 1.55 newtons cancel. Also, of course, we can work out 2.5 newtons minus 1.55 newtons on the left-hand side. Overall then, this leaves us with the following. 0.95 newtons is equal to 𝐹 sub 𝑑. In other words, the magnitude or size of the drag force happens to be 0.95 newtons. And this is exactly what we’ve been trying to find in this question all along, which means that we’ve reached our final answer. The drag force on the bat is 0.95 newtons.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions