A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (d) How much work is done on the crate by the normal force? By gravity?

Video transcript

Hey everybody. So today we're being asked to calculate the work done by both gravitational force and normal force on a box. That is being that has a force applied to it by a child. Now, we have a lot of figures here in this question. We have the box has a weight of 145 newtons. The force applied is parallel to the surface and moving it at a two m, moving at two m at a speed of 0.2 m per second. And that we have a coefficient of kinetic friction. But taking a closer look at the question, we realized that we really don't need a lot of this and I'll explain why right now. So we have a box, right? And it is on a horizontal surface. Oops, let me draw this nicely. So we have our box. Oops, geez we have our box. It has a weight. Well, it has a weight of 145 newtons. So that means that we don't know the actual mass of the box, but we know that the force due to gravity or MG, is equal to 145 newtons. Right? So that is the weight. We also know that therefore if we want to find the work done by normal force, while normal force is a force that acts perpendicular to the um perpendicular to the surface on which an item is placed on. So the normal force would be this way And it would also be equal to MG. So the normal force would also be equal to 145 newtons just in the opposite direction. But let's think about this more critically both the weight and the normal force are acting perpendicular to the horizontal surface. So what does this mean? Well if we write out our definition of um work, we know that work is equal to the force applied multiplied by the distance, multiplied by Kassian data. Where data is the angle formed in case the object is at an incline or the angle formed between the objects path or path of motion and the horizontal between the path of motion. Sorry I was just with my hands and not on screen in case it's at an incline. The path of motion would be at an angle with the horizontal but in this case we are the child, the little baby man is pushing the box against the horizontal axis. So their angle the angle will be zero. So if data is equal to zero then co sign data Is equal to one. So we can effectively rule out the cosine theta. Now taking into factor and I'll write work of weight. The work of weight is equal to MG MG. Multiplied by the distance, distance, multiplied by cosine theta which is zero or sorry, which is one MG times distance. Well, if we're taking into account the work done by weight. Well the weight and the normal force will both cancel each other out right because it is only if weight and normal force cancel out that the object won't sink into the floor or fall upwards, fall upwards, go upwards and move in a horizontal direction. So because weight and normal force cancel each other out, the work done by them therefore will actually be zero for both because they are not doing the work, it is the child's force. It is the child's force that is actually doing work through a distance. The weight and the normal force effectively served to prevent it from falling into the surface and allows it to interact with the surface to get us a force of friction in the opposite direction as well. Therefore we can say that our answer. In this case the work done by both gravitational force and normal force will be zero jewels or answer choice. A. I hope this helps. And I look forward to seeing you all in the next one.