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Cover Page [1,QUESTION 1. 1,1,I] Define average speed and write its S_ L.unil(b)Define projectile motion.(c)State Newton s second law of motion(d) block of mass re...

Question

Cover Page [1,QUESTION 1. 1,1,I] Define average speed and write its S_ L.unil(b)Define projectile motion.(c)State Newton s second law of motion(d) block of mass rests on the table, shown in the diagram: A light rope is attached to it and runs over a pulley: The other end of the rope is attached second block The coefficient of kinetic friction between the block and table is Block exerts a force due to its weight; which causes the system (two blocks and string) to accelerate_ Draw the fre

Cover Page [1, QUESTION 1. 1,1,I] Define average speed and write its S_ L.unil (b) Define projectile motion. (c) State Newton s second law of motion (d) block of mass rests on the table, shown in the diagram: A light rope is attached to it and runs over a pulley: The other end of the rope is attached second block The coefficient of kinetic friction between the block and table is Block exerts a force due to its weight; which causes the system (two blocks and string) to accelerate_ Draw the free body diagram for the block



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Blocks $A, B,$ and $C$ are placed as in Fig. $\mathbf{P} 5 . \mathbf{1 0 3}$ and connected by ropes of negligible mass. Both $A$ and $B$ weigh $25.0 \mathrm{~N}$ each, and the coefficient of kinetic friction between each block and the surface is $0.35 .$ Block $C$ descends with constant velocity. (a) Draw separate free-body diagrams showing the forces acting on $A$ and on $B$. (b) Find the tension in the rope connecting blocks $A$ and $B .$ (c) What is the weight of block $C ?$ (d) If the rope connecting $A$ and $B$ were cut, what would be the acceleration of $C ?$

Haider. So in this problem we need to determine the angle that the strings make with the vertical. So in here we have a dice. And the first thing that we need to do is to draw all the forces that are acting on the dies. And then appliance need to second law we can obtain equations for the Y. And the X. Company. So doing that we find that we have attention that we call team that is due to this string. And we also have the weight of the dice that we call. W. We also note that the system has an acceleration because the dice is inside of a car and it axillary res from from velocity in six seconds. That velocity is that we call final velocity Is equal to 28 m/s in a time of six seconds. So um we first are going to find the equations of for the dynamics of this. So some of the forces in the X company we have that this is detention sign of tita because the angle is with respect to the vertical and this because the car is moving in the horizontal way it is moving in the ads access. So dad we will have an acceleration in this company. So this is mm which is the mask of the dice times the acceleration for the white company of the forces. We will have two forces which are T casino tita in the positive part of wine and minus their weight in this case the dice is not moving in In this company. So it is equal to zero. Now from this we obtained that the tension times the cosine of theta is equal to the weight. We know that the weight is equal to the mass times the acceleration due to gravity. With these two equations, we're going to call this one in these two we are going to determine the angle tita and what we can do is to eliminate detention in the other equation so that we can divide want over to Yeah, Equation 1/2. So that we could see that these will give us deal sign data over T of consign of tita and for the right hand side we will have and A times A. M. Of G. So we have that. We can eliminate this too. And we also eliminate the tension so that we know that sign of tita. Over a dozen of teacher is the tangent of tita. And then we will have a over T. So to obtain the angle tita, we take the inverse function of the tangent function for both of the size of these equations. So we won't have Tangent of -1. All of the acceleration over the acceleration due to gravity. We know the acceleration due to gravity because this is the value of 9.8 m/s. But we don't know the acceleration of the dies or the car in which the diet is in it. Sm So we need to determine the acceleration but we know from the beginning that we are given a velocity and the time. So you think any memories? Yes. In key markets we can solve this by using the following equation. We know that the final velocity is equal to the initial velocity plus the acceleration times at time T. And we know that the initial velocity is zero and the final velocity is the one that we are given. So solving for a, we know that this is the final velocity over the time. We know that the final velocity is 28 meters per second and the time is six seconds. So from here we obtain about you off. Yeah. Mhm. For 0.6, see seven meters per second is square. So now we know the body of the acceleration. So we just simply need to substitute that value into the equation for tita. And then we can obtain the value for the angle tita. So it does regions substitute those values. The acceleration is 4.66 m/s square and the acceleration due to gravity is 99.8 m/s square. So good in this, plugging this into the calculator, we obtain about you. Yeah, 25 point 48 degrees. And this is the angle the district make with the particle

So here we control all the system. It's is this a frictionless massless pulley? Aah! This would be, ah, maths. A going up would be the force normal of a going down would be, of course, um sub a g and going to the right butts. This would be massive. Be going to the right here would be forced. Tension going up here would be forced. Tension going down would be m sabi tree and this is the full free body diagram for the system. Now we can say that the force tension is going to be equal to rather this would actually my apologies. This would be your answer for part A. So just party just asked us to draw the free body diagram. This would be the full free body diagram for part B. However, now we need to solve for the acceleration. So here. Ah, for block A. There is no motion in the vertical direction. Therefore forced normal of a wood equal the mass of a g again. This plane right here. This table is perfectly horizontal. Therefore, the normal force of block a on block a rather would be equal to the weight of block A We can apply Newton's second long to block a in the ex direction. This would equal force tension and this would equal the mass sub a times acceleration of a in the X direction. We can say that the sum of forces for block B in the UAE direction would be equal to M C B G minus forced Sebti Aah! This would equal and sub B times a. So be in the wind direction. Now the two blocks air connected. Therefore, the acceleration of block A in the ex direction with the equal to the acceleration of block B in the wind direction. And we're going to call this simply a So we're gonna lose the sub scripts for a and then we can combine the two force equations. We can essentially combine these two equations to solve for a so we can say that forced tension would then be equal to m sub a time I'm so be she with minus force sub t would be equal to m sub b a and then we're going to say that m c b g minus m c a a cool m sub b times a uh this would become massive eight times today plus massive B times A would equal mass Sabi G And we can say that the acceleration would then be equal to the massive be times G divided by the sum of the masses. And this would be your answer for the acceleration. At this point, we want to find force tension. We know forced tension would be equal to again m sub a times, eh? Therefore, this would be equal to g times m sub a m sabi divided by again the sum of the masses m sub A plus I'm sabi So this would be your force tension here That is the end of the solution. Thank you for watching.

Free by diagrams party in the first block over here on the surface. So you have normal force going upwards. You have mass times grab me the weight going down and then you have the force attention being pulled on. Then for second block, you have force of tension being pulled upwards and then you have mass two times gravity weight being pulled down. As for the free body doctors since two blocks are connected, the maxims of their accelerations will be the same. So what a one. The extraction equal a to in the y direction the T equals the overall acceleration. So if we combine forces equations in second law, we get this is on for part de now. So force attention equals and one times a that's two times gravity. My eyes force attention equals mass to times acceleration. So rearrange subs to you get massive too Times gravity, my eyes, massive block One times acceleration equals mass to times acceleration, the neck business and to times a being and one times a plus and two times a equals and to times she so you re range and solve for a celebration equals gravity times. That's what to over and one plus into forced tension came equals mass and one times a equals g times m one and to over and one plus. And to this is for your part a free, wide diagrams over here. And then we saw four acceleration terms. Love Newton's second law, all right?

Party of this problem is asking to draw free body diagrams for both Block a lot B for Block A. We have a normal force upward. We have a attention forced the right. I'm getting to know my team won. We have the weight of a downward and we have the first in force on A to the left for Block B, which is sitting on the ramp. We have a tension force. Words t two. It's different than t one. We also have two U one downward. Now, this is the same as this team won because they're connected to each other. We have the normal force of being perpendicular to the ramp. You have the friction force of B opposing the motion down here, and we have the weight down here like this. Of course, we can split this up into its components like this. This becomes the Baby Terms co signed off thirty six point nine degrees because that's what this angle here is. And this becomes w time sign of thirty six point nine. And I believe that's answered a party there. I think I've got all forces for part B. We want to figure out what Team one is the tension between Block's A and block speak. So what we do those were going first, Teo the sum of the forces in the wind direction, which is equal to zero. Since there's no acceleration wind direction, we're going to do it on a and just give a story. That's a that's B. This implies that the way today equals the normal force because they're the only sea forces in the white direction and they're in the opposite directions Now. We're gonna do the sum of the forces in the extraction on eh? And again secret zero. Because although they're moving, they're not accelerating. And so Newton's second law there gives us that to you. One is equal to the friction force on a but the friction force on a can be expressed as the coefficient of connect friction times, the normal force. You found the normal force here equal to the weight of a Now these are both given in the problem, and so we can easily calculate this to be eight point seven five moons. And that's attention in the road between Andy now in part. See, we want to figure out what the weight of block C is. And so the forced I grand for Block C looks like this. We have tension t to t, too, because it's connected to block too. Going up like that and we have the way to see pulling it down. Newton's second law, some of the forces and why equals zero because there's no acceleration implies that attention are up to is equal to the way they see. And so if we can figure out what T too is, we'LL have the answer because we're trying to solve for the way to see, so to figure out t two we're gonna look at Block B and we're going to take the sum of the forces long along the ramp equals zero because there's no acceleration along the ramp. This is again for Block B gives negative friction. Forces B is playing in the negative direction. Mine s t one minus the way to block B times Sign of thirty six point nine plus tea too is equal to zero solving forty two. This gives T, too is equal to t one plus way to be time sign thirty six point nine plus version force of B Now we want to figure out what this friction force, if he is. In order to do that, we're going to do the sum of the forces perpendicular to the ramp. It's equal to zero because there's no acceleration in that direction. And this equation implies that the normal force of like B is equal to the way Toby Times go sign of thirty six point nine. Well, we know that the weight of Block B is so we can usually calculate the normal force here and people twenty Nunes. This implies that the friction force of B is equal to zero point three five, which is just the coefficient of connect friction. They gave us times normal force and this gives us seven Nunes. So now we know T when we calculated in part a ueno the way to block B. This is just sign of thirty six point nine, and we just calculated first enforce a B. So you Khun solve easily for t two. And when we do that, we get thirty point seven six Nunes again. This implies that the weight of C is equal to thirty point seven six Nunes, and that's answer for Park si now for part D were cutting the rope between AM B and we're trying to find the acceleration of block C. So now if I re drawn my force diagram for Block B, I attention t two moving up like that, I still have my normal force in this direction. I still have a friction force here and I still have the wayto be pulling a vertically down. I kind of failed drawing it there. But it's going vertically down now because the angle here doesn't change. The way doesn't change. That means the normal force is not going to change, which also means that the friction force is not going to change. So it can bring over the fact that the friction force on B is equal to seven Nunes that still valid here. The only thing that changes is there's not a tension pulling left here, so there will be an acceleration to the right. So the way we actually saw that is we do some of the forces along the ramp. In this case, it's not equal zero. It's going to be equal to the mass times acceleration along a ramp. But the sum of the forces just gives us tea, too. Minus friction on Block B minus way, Toby Time sign of thirty six point nine. And again, it's not Equals zero and secrets of the massive block B times The Acceleration Block B by noon, second law. Now the mass of Block B is equal to the weight of Block B, divided by G, so they can plug in both these values and we get to point five five kilograms. And so we have two point five five here. And if we're solving for a B, we just need to calculate this and then divide by two point five five. No, everything is known at this point. And so when we solve, we get an acceleration in Block B three point four three minutes for second square. And the key idea now is that the selection of Block B is tethered to the solution. Glossy, they're going to be the same. We're going to celebrate at the same rate and so they Acceleration of sea is also three point four three meters per second squared, and that's the final answer for party


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