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Juana slides a crate along the floor of the moving van. The coefficient of kinetic friction between the crate and the van floor is $0.120 .$ The crate has a mass of $56.8 \mathrm{kg}$ and Juana pushes with a horizontal force of $124 \mathrm{N}$. If $74.4 \mathrm{J}$ of total work are done on the crate, how far along the van floor does it move?

$1.30 \text{ m}$

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So this question gives us a bunch of different other information and then asks us to figure out how far a crate moves given the total work horizontal force coefficient, friction and a mass of the crate. So I started with a diagram here. But let's start by extracting all the rest the information from the problem and see if we can come up with a plan of how to solve this. The first pieces of information given in the problem are the coefficient of friction between the create in the van, for which is 0.1 to 0, the mass of the crate, which is 56 0.8 kilograms. And the fact that we've got an applied for us, which I'm gonna label for sap of 124 Newtons and the total work done on the crate as 70 four point for Jules here. And they want to know how far along the van floor the create move. So we're gonna call that displacement, and that's ultimately what we're trying to find here. So one of the key words I want to emphasize here is the fact that they ask for the total or rather, they give us the total work. The difference here being some questions sometimes ask how much work does a single person do, or a single force doing an object? But because we're looking for the total work, If we look at our force formula, work equals force, times displacement. If we're asked to solve for the total work, we're gonna have to sell for the total force or a net force in order to figure out the overall displacement. So that's kind of the first step to this question. Can we use the information given in his problem to find Net Force? Because looking at this formula, we have the work. We're trying to find displacement. So what we're gonna have to calculate here is for us again. The temptation might be to just use the 124 news that we're told is the Applied force. However, the issue with this is that is the applied for us on Lee by the person pushing the crate in neglects to include the force of friction that is opposing the crate, which is the next thing that we're going to talk about and so for. So we have to solve for the net force. So to do this, we go back to our net force concepts which will be analyzing that create and drawing a force body diagram on this. So the forces that are acting on the creed in the X direction we have the force applied. I'm assuming is going this way, working against the motion. We have the force of friction going the other way and then in the Y direction there's no motion. But we do have two forces that will have to include You've got the normal force and we've got gravity. So because their motion is in the X direction, that's the direction that I'm gonna care about for finding my net force. Now. I already know the force applied, which is great. So in order to find the net force in the extraction, the only unknown force I have is FF That's right, that I would explicitly so f net in. The X direction is equal to FAA in the positive X direction and subtracting the force of friction because it's pointing the other direction again. We have f A So all I need to calculate is this FF piece to do that you might recall that the formula for force of friction is force of friction equals the coefficient of friction times the normal force in this case, because the normal force and gravity of the only two forces in the why and there's no motion in the Y direction, normal force is exactly equal to gravity. So I can just replace this with them. You f g and then using the formula for gravity, replace that F G within energy. So plugging that in for my frictional force, I get my net for us in the X direction equals F A minus meu m cheap Now, looking at the information that we have on the left hand side here we again No RFK, where given them you were given the mass of the create and were given gravity. So now I have enough information to solve for my net force in the X direction, which means I have enough information to solve for the displacement. So let's go back to that work formula. So work again was equal to force times displacement. Solving this equation for displacements I get displacement is work over force again. That's the net force plugging in my answer for net force here I get that my displacement is equal to the total work done over the Applied Force minus mu M g. So l directly this would be my final answer. And again, we're given all those values on the left hand side here and in the question. So plugging those all into a calculator, I get around ID to appropriate significant digits 1.30 meters. Make sure that when you're punching it into your calculator, depending on the calculator that using you put the entire denominator in brackets or you will not get the

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