A rubber-wheeled 50 kg cart rolls down a 15 degrees concrete incline. What is the magnitude of the cart's acceleration if rolling friction is (a) neglected and (b) included?

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Hey, everyone in this problem, a 30 kg ice lab slides on a snowy plane inclined at 18 degrees with the horizontal. We're asked to determine the acceleration of the ice lap. If there is friction on the snowy plane, we're told to consider Mu K to be equal to 0. for ice on ice. We're given four answer choices. Option A negative 2.75 m per second squared. Option B negative 3.75 m per second squared. Option C negative 2.9 m per second squared and option D negative 3.9 m per second squared. Now we're gonna start by just drawing out what we have. So we have this inclined plane that makes an angle of 18 degrees with the horizontal and we have our ice lab, we're just gonna draw as a rectangle that's gonna be sliding down that incline. Now, because of gravity, if we leave it alone, that ice lab is gonna slide downwards on that incline. That means that friction is going to act up that incline because it opposes the motion. All right. So we're asked to find the acceleration. Let's go ahead and draw a free body diagram. And we know that through Newton's second law, we can relate the acceleration to the forces. As we draw free body diagram, take a look at the forces and see whether we can use that to find our acceleration. So we have our free body diagram, we have our snowy ice slab. And what forces do we have? Well, I think upwards, ok, perpendicular to the inclined plane is going to be the normal force acting to the right parallel to the incline plane is gonna be the force of kinetic friction F K and acting straight down vertically is the force of gravity. Now, what we wanna do is take a tilted axis. So we want our X axis to be parallel to the incline, we want our Y axis to be perpendicular to the incline. And so we need to break this force of gravity up into components X and Y components for this tilted axis. So we do that pointing to the left, we're gonna have the X component of the force of gravity F G X and pointing down perpendicular to that plane, we have the force of gravity in the Y direction or the Y can point OK. Now, the angle between the force of gravity and the Y component is going to be degrees based on our 18 degree incline. Now, if we look at our answer choices, we can see that they're all negative So what we wanna do is we wanna choose our frame of reference so that we get a negative acceleration in this case, OK. We can choose our frame of reference either way. But in this case, we have an acceleration that's negative, we have a block sliding down because of gravity, but being opposed due to friction. So we're gonna take our frame of reference to be up into the right now when I say up into the right, I mean, relative to our tilted axis. So let's get started with Newton's second law, Newton's second law tells us that the sum of the forces and in this case, we're looking in the X direction, it's gonna be equal to the mass multiplied by the acceleration in the X direction. Now there's some of the forces in the X direction. On the left hand side, we have the force of gravity in the X direction and that's acting in the negative X direction. So we have negative F G X and on the right hand side, we have the force of kinetic friction that's acting in the positive direction. And so we're gonna add F K and that's are, those are all of the forces acting in that XX component in the X direction of our tilted axis. This is gonna be equal to the mass multiplied by the acceleration in the X direction. Now that X component of the acceleration, that's what we're looking for. OK. That's the acceleration of the block because that's the motion direction of motion. OK. It's gonna, all of its motion is gonna be in that tilted x axis. All right. So this force of gravity, the X component, how do we break this up? Well, this is gonna be related to the angle through sign. OK. Because it's on the opposite side. So we're gonna have negative sign of degrees multiplied by the force of gravity which is the mass multiplied by the gravitational acceleration. G. What about the force of kinetic friction? F K? Well, this is gonna be equal to mu K multiplied by the normal force. N so we have negative sign of degrees multiplied by M multiplied by G plus mu K multiplied by the normal is equal to the mass multiplied by the acceleration in the X direction. Now we know the mass, we know the gravitational acceleration, we wanna find the acceleration A X but we don't know the normal force. So how can we find the normal force? Well, the normal force acts in our tilted Y axis. OK. So we need to look at the sum of the forces in the Y direction as well. Now, in the Y direction, this block is not moving. OK. It's not moving up and down. So it's an equilibrium which means that the sum of the forces is equal to zero. What forces do we have? Well, we have the normal force acting in the positive Y direction. We have the Y component of the force of gravity acting in the negative component Y direction. Sorry. So we have N minus F G Y is equal to zero. That tells us that the normal force is equal to the F or the Y component of the force of gravity F G Y. Now the Y component is gonna be related through the cosine of the angle because we're talking about the adjacent side. So we get cosine of 18 degrees multiplied by the mass multiplied by the gravitational acceleration G. OK. So now that we've figured out our normal force, we can get back to our, some of the forces in the X direction so that we can solve for that acceleration. So we have negative sign of 18 degrees multiplied by M multiplied by G plus UK multiplied by cosine of degrees multiplied by M multiplied by G is equal to mass. M multiplied by the acceleration A X. Now, every single one of these terms has a mass in it. So we can divide the left and the right side by the mass and those are all gonna go away. And so we have our acceleration A X, it's gonna be equal to everything on the left hand side here. And we're gonna substitute in those values. So we have negative sign of 18 degrees multiplied by 9.8 m per second squared plus mu K which is 0. multiplied by cosine of 18 degrees multiplied by 9.8 m per second squared. And if we work this out on our calculator, we're gonna get an acceleration value of negative 2.75 m per second squared. All right. So we've made it to the end. We've determined that the acceleration of the ice lap, considering this friction is negative 2. m per second squared, which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

A rubber-wheeled 50 kg cart rolls down a 15 degrees concrete incline. What is the magnitude of the cart's acceleration if rolling friction is (a) neglected and (b) included?

Verified Solution

Key Concepts

Newton's Second Law of Motion

Gravitational Force Component

Rolling Friction