A baggage handler drops your 10 kg suitcase onto a conveyor belt running at 2.0 m/s. The materials are such that μₛ = 0.50 and μₖ = 0.30. How far is your suitcase dragged before it is riding smoothly on the belt?

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Hi, everyone in this practice problem. We're being asked to determine the acceleration and also the time required by a crate to come to reach the bottom of a ramp. We have a crate with a mass of 25 kg placed on a ramp inclined at an angle of 30 degrees with the horizontal. The creed has a coefficient of static friction with the ramp of 0.4 and a coefficient of kinetic friction of 0.2. The ramp has a length of 10 m and we're being asked to determine the acceleration of the crate down the ramp and the time required by the grade to come to reach the bottom of the ram. Considering the direction pointing up the ram is positive. The options given are a, a equals 1.9 m per second. Squared, T equals 2.5 seconds. B A equals negative 3.2 m per second squared, T equals 3.43 seconds. C A equals 1.9 m per second. Squared T equals 3.43 seconds. D A equals negative 3.2 m per second squared T equals 2. seconds. So we will consider the crate as a particle. And I am going to start us off with drawing a pictorial representation of our system. So first I'm going to draw the ramp, this is going to be the ram which will make an angle of 30 degrees with the horizontal. And then uh I'm going to still indicate the crate as just this square right here. But essentially this is just a particle. So the crate will experience multiple different forces. The first one since it is going downhill, so we know that it will have a frictional force pointing the other way or the other direction from which the crate is going downwards. So FK is pointing upwards and it will have a gravitational acceleration acting on the um crate. So it will have an FG or a gravitational force which will just equals to M multiplied by G, it will then have a normal force acting just like so and this will be uh represented by this and right here. So I am going to um just uh put a convention at which the blue marking here is the X direction and the Y direction is going to be a little tilted just like. So to make it easier for us to actually do this problem. So it is given in the problem statement that us is 0.4 and UK is 0.2. And at the same time, I'm going to just start us off with doing the uh projection for the FG, we know that the ram is 30 degrees. So we know that the angle here, it's also going to be 30 degrees just like. So, and therefore, in the X direction in the X direction, we will have FG multiplied by sine of 30 degrees. And in the uh in the uh wide direction, we will have FG multiplied by cosine of 30 degrees. Awesome. So now let us actually consider the X axis along the incline plane and the Y axis perpendicular to the X axis, there will not be any acceleration in the Y direction. So therefore, by Newton's first law, Sigma Fy will be equals to zero. So then we know that in the Y axis there will be N and only FG multiplied by cosine of degrees acting on the Y axis which will be equal to zero. So then N will be equal to FG multiplied by cosine of 30 degrees which will be equal to M multiplied by G multiplied by a cosine of 30 degrees which we will um substitute all of the values in M is 25 kg. Given in the problem statement, G is just 9.81 m per second squared and cosine of 30 degrees is just cosine of 30 degrees. And calculating this, the normal force will come out to a value of 212.176 Newton. Now for the friction force FK, we know that um FK will be equal to UK multiplied by N. So UK is given to be 0.2 and the N is 212.176 Newton, which we have just previously calculated and calculating this, we will get the FK or the kinetic friction force to be 42.435 Newton. Next, we want to look into the X direction or the X axis where sigma FX is not going to be equal to zero, but instead it is going to be equal to M multiplied by A X. So using Newton's second law in the X direction, there will be the FG sine 30 degrees which is the projection of the gravitational force and the kinetic friction. So um using uh the convention listed in the problem statement where the direction pointing up the M is positive. So we know that FK, in this case, minus the, the FGS sign 30 degrees will be equals to M multiplied by A X. So in this case, we will then know that or we can then calculate that A X, which is what we want to obtain from the um Newton's second law can be obtained by FK minus FG multiplied by a sign of 30 degrees divided by M. Let's substitute the value FK is 42.435 Newton minus FG multiplied by a sine 30 degrees FG is just M multiplied by G So that's going to be 25 kg multiplied by 9.81 m per second squared, multiplied by a sine of 30 degrees to fight that with mass, which is going to be 25 kg. And then calculating all of this, we will then obtain the A X or the uh acceleration in the X direction. Didn't be equal to negative 3.2 m per second squared just like. So next, what we wanna do is to use the kinematic equation where delta X will be equal to P knot multiplied by T plus half A T squared or A XT squared. So in this case, we know that the delta X will be equals to negative 10 m. And then the A um P knot is going to be equals to 0 m per second. And the acceleration is just what we obtained previously. What we are interested to find is the time component because that is the remaining thing that we need to still calculate. So then let's substitute off our information. Delta X is negative 10 m equals zero plus half, multiplied by negative 3.2 m per second squared, multiplied by T squared. And then we then know that T squared will then be equal to 6.25 seconds squared after some calculation and rearrangement. So then um taking the square root P will then come out to a value of the square root of 6. seconds squared, which will be equal to 2. seconds. So there we have the time required which is 2.50 seconds and the acceleration which is negative 3.2 m per second squared, which will the combination of both things will then correspond to option D in our answer choices. So option D will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that'll be it for this video and thank you so much for watching. Have a good one. Bye bye.

A baggage handler drops your 10 kg suitcase onto a conveyor belt running at 2.0 m/s. The materials are such that μₛ = 0.50 and μₖ = 0.30. How far is your suitcase dragged before it is riding smoothly on the belt?

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Key Concepts

Friction

Newton's Second Law of Motion

Kinematics