 In this problem, we have two boxes sitting on the surface. They're currently sliding to the right and they're also pushed to right by a push of 20 Newtons and the question is what is their common acceleration to the right? I've already prepared the whole setup for the problem here. First, I'm going to draw the free body diagrams of both objects then I'm going to look for additional equations and At last, I'm going to plug in Newton's first and second law for both objects in X and Y direction Let's start with the free body diagram of box B I isolate box B from the drawing and wherever I cut through some force I'm going to replace this By a force in the free body diagram To start with on the left side, I was cutting through a surface. So there must be a normal force from the left to towards the right It is actually the force of A on B. I do have another surface, which is the one Below which will also cause me normal force going away from the surface. So I have a normal force on B from the surface And it will also cause me a friction. The friction always Opposes the possible sliding. So if I'm moving to the right, that means my friction On B will be to the left I'm on the surface of the planet. That means there also must be a gravity Next is the free body diagram on A I'm doing the same thing isolating it from the environment And this time from the left I'm having a push From the bottom, I also have a normal force And the friction same as before the friction will try to oppose the possible movement From the right side, I have a force also going to the left Which will be the force of B on A Which is a Newton's third law pair with the force on B by A And once more, I'm having gravity So I have to draw a gravity on A In both cases, I'm going to assume the acceleration is to the right I might be wrong if I am then math, which is give me a negative Value for the acceleration So it's named this acceleration A and the other one is acceleration B For the coordinate system of A, I'm just going to use a normal coordinate system with x to the right And y up The same for B I'm going to use x to the right And y up So once our free pi diagrams are done, I'm going to look for additional equations I do know that my force of gravity on A Is equal to mass of A times g So 100 Newtons I also know that my friction on A I better label this as the free pi diagram as friction on A And the magnitude will be equal to my coefficient of kinetic friction times the normal force on A Similar for B. I do know that my force of gravity on B Is mass of B Times gravity Which will be 20 Newtons, and I know that my friction on B Is going to be mu k times The normal force on B I also have some bridge equations between the two. For example, I do know that the force On A by B The magnitude of it is equal to the force by B on A And so I'm just going to call this force in the future Also, I do know that my two accelerations must be the same As they're touching each other. That means the two magnitude are the same So I'm just going to call this acceleration Now I'm ready to look at Newton's first and second law for both objects In their x and y directions For object A in x direction, I'm clearly having Newton's second law As it is moving to the right and potentially accelerating to the right So some of our forces In x direction equals mass of A Times the acceleration In y direction, I'm not accelerating. So I have Newton's first law some of our forces in y Equals zero And for box B, I'm having exactly the same thing Some of our forces in x equals mass B times acceleration And in y direction the sum of our forces in y Zero Now I'm going to go through each of my free body diagrams and look at which direction The force is acting on if I go in my free body diagram of A. I have a force a push I have Friction and I have a force In x direction So I'm going to put them in my x direction. So I have A push Plus a friction on A plus Force on A by B, which I just called the force And that's in x direction. I'm having an acceleration in positive x direction At least that's what I assume And I have all my forces in x In y direction. I have a normal force And I have gravity So I have a normal force Plus gravity Zero Next I'm going to look at the direction of each of those forces in x. My push is in positive x directions I am writing plus the push My friction is against my x direction. So minus friction S is my force a B of which the magnitude I called f before Is equal to as my acceleration is positive and writing m a times acceleration I know the push. I do not know the friction and the force from B on A. So I cannot further solve this equation So I'm going to go over to the y direction Where my normal force is in positive direction My gravity is in negative direction So I do know that my normal force Is equal to my gravity Which I already calculated is 100 newton If I look at my additional equations up here This gives me directly Another value namely the friction I say it's the coefficient of friction times the normal force Which is 20 newtons. So I can plug this in Back In my x direction. So I'm getting Plus 20 newtons from my push Minus 20 newtons from my friction Minus the force on A by B is m a That is a It's interestingly the push is just cancelled out by the friction So I get that my acceleration or is Minus whatever B is doing on A over to mass on A Now let's see what we get for object B For object B In x direction I have My force of A on B And I have my friction So I have my force Of A on B Plus my friction on B Is mass B times acceleration B In y direction I have gravity and the normal force Which must be equal to zero So a bit a I'm now looking at the directions my force of A on B is in positive direction. So positive my magnitude F My friction is a negative direction And my acceleration. I'm also steering positive For my y direction I get plus the normal force Minus f g It's actually f g of B. So I should write the subscripts Zero so as before I can easily get my normal force of g B Which in this case Is 20 Newtons Once again, I can use the equation for the friction which will give me four newtons Plugging this back in I get plus my fraction Yes, sorry plus my force from A on B minus my four newtons Is mass of B times acceleration As I want to solve for the acceleration I'm gonna solve this for the Force between A and B which I'm not needing m B times acceleration plus four And then I'm substituting this in My equation that I already got here So I have my acceleration Is equal to Minus Mass of B times acceleration Plus Four divided by the mass of A So I have mass of A times acceleration So 10 a Is equal to minus Mass B times a so to a minus the four So 12 A Is equal to four Therefore my acceleration is minus 0.33 meters per second square This means That the acceleration that assumed to be to the right is actually not to the right but to the left That means although there is a push from the left My system will eventually be slowing down. It starts at an initial velocity of 10 meters per second to the right But will eventually slowing down. So I'm gonna give the answer as a vector. Therefore my A as a vector is 0.33 meters per second square to The left Let's say it's even slowing down and this will be my final answer