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7.99 - CHALLENGE PROBLEMSPoints:5Shown below is a 34.0–kg crate that is pushed at constant velocity a distance 8.5 m along a 38° incline by the horizontal...

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7.99 - CHALLENGE PROBLEMSPoints:5Shown below is a 34.0–kg crate that is pushed at constant velocity a distance 8.5 m along a 38° incline by the horizontal force [(F)vec]. The coefficient of kinetic friction between the crate and the incline is μk = 0.384. Calculate the work done by the applied force. Tries 0/5 Calculate the work done by the frictional force, Tries 0/5 Calculate the work done by the gravitational force. Tries 0/5 Calculate the work done by the net force.

7.99 - CHALLENGE PROBLEMSPoints:5Shown below is a 34.0–kg crate that is pushed at constant velocity a distance 8.5 m along a 38° incline by the horizontal force [(F)vec]. The coefficient of kinetic friction between the crate and the incline is μk = 0.384. Calculate the work done by the applied force. Tries 0/5 Calculate the work done by the frictional force, Tries 0/5 Calculate the work done by the gravitational force. Tries 0/5 Calculate the work done by the net force.



Answers

A $15.0$ -kg block is dragged over a rough, horizontal surface by a $70.0$ - $\mathrm{N}$ force acting at $20.0^{\circ}$ above the horizontal. The block is displaced $5.00 \mathrm{~m}$, and the coefficient of kinetic friction is $0.300$. Find the work done on the block by (a) the 70 - $\mathrm{N}$ force, (b) the normal force, and (c) the gravitational force. (d) What is the increase in internal energy of the block-surface system due to friction? (e) Find the total change in the block's kinetic energy.

So we have a 10 kilogram uh, create a nen coin like this and there's an applied force pulling up the ramp of 100 Newtons. Um, we also know that the coefficient of friction is 0.4, and the initial velocity is 1.5 meters per second, and this total distance of the ramp is five meters. Oh, and this is 20 degrees. That's okay. So that's pretty much everything we're given now. First, want to find the work done by gravity, and they're a couple ways to do this. I'm going to say that it's going to be the negative change in gravitational potential. And that way I could just do with potential. I don't have to worry about any forces or distance or anything like that. So this is gonna be minus, um, MG times. Uh, well, Delta. Why? So why final minus. Why initial and why Initial is zero. So this is just wife final, so that's minus 10 times 9.8, and then the wife final will we know the high pot news That means the height of the ramp. It's gonna be five meters. The happened News times the sign of three angle Fada. So 20 degrees. And this gives us 100 minus 168 jewels of work. It's important that it's minus because gravity is doing negative work. It's the forces pointing down, but the crate is going up. Ah, And now we want to find the increase in internal energy due to friction. That'll be we'll call that Delta E. And again, I can do a similar thing to what I did for the gravity. I can say the work due to friction is minus changing energy. And then the work due to friction. Well, that's gonna be, um, the force of friction, which is mute times the normal forced times the distance that it travels. And since friction is pointing in the opposite direction to the blocks are the creates motion. The work is actually gonna be negative. That means both the negatives cancel out now, the normal force, Uh, you haven't memorized. It's easy to know what the normal force will be for something like this, but we just, um let me draw a quick, free body diagram like this. So we have gravity pointing down this way and the normal force here. So we know that the normal force and the the component of gravity that's perpendicular to the ramp or the surface of the creed sliding on are going to sum to zero because it's not moving in that perpendicular directions. This will be minus mg and then co sign theater because we're going perpendicular and that has to equal. Zero is the forces are balanced. That means the normal force is mg coastline Fada. So I can plug in here again. Same you mg co sine theta times The distance is the change in energy. Um, you is 0.4. The mass is 10 kilograms G's nine point h take co sign of the angle 20 degrees and then multiply that by the distance, which is just five in this case because we're looking at the length of the ramp, Not not the height, Um, and you multiply all that out and you get 100 84 jewels. Um, that's the changing energy or the increase in energy due to friction. Um, so now we want to figure out the work done by the Applied force of 100 Newtons. So it's just gonna be the force times the distance. Since there parallel to each other. That's 100 Newtons. Times five meters comes out to 500 jewels of work. And now the change in kinetic energy. Uh, that's going to be the sum of all the works that we've found. So we have the work applied, and we also had worked due to friction. They also had work due to gravity. This is gonna be 500. Um, the work due to friction. Well, here we got 100 84 jewels, but that was the changing energy. And remember, we canceled. The negatives of the work due to friction is actually negative. Uh, 1 84 No work due to gravity is also negative. It's 168 and is ah, 148 jewels. So now that we know the change in kinetic energy, we can figure out the speed of the crate because this is gonna equal, um, 1/2 m. I say the final squared minus the initial squared, which is the total work. So v initial script we were given and the problem I run appear, it's 1.5 meters per second. Okay, so this is gonna be, um we're solving for V final here. So, uh, say the final square, this is gonna be too terms the work divided by the mass plus the initial velocity squared will take the square root of all this and then plug in the numbers that we have. The total work which we found in the last part, is also the change in kinetic energy 1 48 divided by the mass of 10 kilograms and add in 1.5 squared and this comes out to 5.64 meters per second.

Hello, friends. This is the problem based on Newton's second up here it is given Ah, block off. Must 40 kg has to be pushed up along the rough inclined plane having inclination 30 degree and coefficient of friction point food bye. Or gentle force Force elf? No. So that it can move up alone. Inclined plane with the cost status feed we have to calculate Worked on first but have played Foot's Yeah, second by friction seen by bait or gravitational force and deep by Netflix. Let us hear various forces acting on it. Date of the block. Empty. Original Applied Force. Yeah, I don't know. No normal reaction. Yeah, yeah. Uh huh. No, let us see it. Yeah. Net force in the origin direction. Well, forces what we have to applied. That is a friction, but which is mute into end. Mhm. So f minus, um UK and cause off 30 minus and sign off. 30 Must be gentle and and cause off 30 minus, um ukip and sign off. 30 minus energy must be equal to zero. So by solving these two equation, we can say f minus road three by two mieux que plus half and to end and e road three by tau minus half milky into l minus mg scotto zero. So we can see. Yeah, yeah. After ability. By solving these two equations, value off f, you will get 4 98 point 168 Newton On a normal reaction, you will get 5 88.566 Newton. So friction you will get UK into end. That is quite foreign. Toe 5 88.566 That is 2. 35 0.43 Newton. Yeah, Bob. Done by Applied Force. Yeah. Applied forces in our gentle direction. So full 98 point 168 I kept dot Displacement has to be resolved into its company. Eight cost 30. I kept bless. 18. 30 Jacob on solving it. You will get 3000 451 point photo. You're b but worked on by friction. Yeah, Friction is 2. 35.43 and 28 cause off quality. So it is 18 83.41 Jordan Seper worked on by weight Anji MG daughter T Yeah, M g is 40 and to 9.8 and to g eat cause off 30. I kept less. Eight. Sign off 30. Yeah, Jacobs history. Negative. 15682 Deep work done by Ned Foot's F Net zero since. Well, the city is constant, that is. Acceleration is zero. So, Bob Denbigh Ned Force is you, That's all. Thanks for watching it.

I have driven. This is the problem based on broken by the forts. Here it is given a thread off must 10 kg is to be pulled over the inclined plane. Having that inclination 20 degree which is to be loved and kinetic friction is find for it is to be pulled with constant speed off. Well 0.5 m due to pulling force off 100. You can if it moves, find meter Then we have to calculate Worked on in the first part Broken by gravitation foods changing Internet energy Toto Fiction she worked in by putting foods change in kinetic energy when it moves 5 m band and took a track And whether city, when it is moving 5 m and that is to start solving it first back Yeah, Welcome. Bye. Gravitational force will be M G s cause off 90 Placed it up. Substitute the volume 10 in 29 point age in tow. 5 m because off 90 plus 20. So it is to be minus 1 68. Dude, this is the answer off part A not part we worked on by friction. Friction will be quarto um UK into normal reaction. She's a normal reaction weight will act here. Normal reaction will be in this direction. You can result a bit empty in tow component. So here it will be empty cost. It'll on Dhere It will be and design oxygen so normally except will be equal toe empty capacitor. So friction will be mu cake and d course it'll Yeah, mhm! So walked on by friction Bilby It's got to change in internal energy worked on by friction That is hell in tow Muche it mg cost itto Yeah, five mil k is 50.4 10 into 9.8 and toe cause off 20. So it is Toby 1 84. Do see part work done by pulling force f l cause off Europe that we correct. Here we are taking distance as forces given 100. Even so, it will be 500 to by barking. Largest serum changing kind of technology is Curtis emission off work minus changing intern energy Applied Burke energy to Europe worked on by force work done by gravity minus changing in turn, energy broken by force 500 1 68 minus Vanity four circuit is Toby 1 48 to e part um changing kind of technology is half and be final velocity minus half and v A script. Yeah. Substitute the value 1 48. Okay, this we're finding can in tow. Initial velocity. That is one point fight. So conserving it. Mhm. Yeah. No final velocity. You will get 5.65 meter per second. Yeah, they're so thanks for

So let us draw the free body diagram first. So this is the inclined plane, this is the horizontal floor and this is the block. This is the block. And in this direction there's a force MG. Scientific to in this accent there's our gravitational force MG. And in this election they're supposed MG cost to to And this will be the normal course on the block. And and in this direction in this direction There's a 4th app and we lost tv. Now this is MG cost tutor. This is M. G. This is N. G. Scientist to this angle of inclination is to to No. In 1st part. In first part we are asked to find out the force acting on the block in so the force will be F. Is equal to MG saying to done in the center from the free body diagram. This party of this problem. So So this will be mass given is 1.25 Casey. Excellent. Send you to gravity is 9.81 While car was signed 30 2010, it is 45° or 30°.. So let us substitute the value Not here. The angle has given us 15 degree. It is 15° As it was 15°.. So after calculation we will get a value of 3.17 newtons. So this much amount of fourth will be acting on the block. So in part B let's move to party there. We need to calculate The work done by the force in order to move a block 3.6 metre of the incline. So in this case the work done will be force multiplier. Where distance table. The force we have calculated as 1.3.17 multiplied by this transferable is given a 0.6 m. It will give us a value of 1.9 jewels 1.9 Julian. So this is the amount of work required to burden. In order to move the block. 0.6 metre update a ramp or the inclined plane. So let's move to part C. In part C. We need to calculate the work done by the gravity for the same part. For the work done by gravity Will be equals two will be false to minus MG. At It will be -1G delta X. Sign. Scientific to their access that distance travel along the inclined plane. So let us substitute the value 1.25 multi pair of Axler center to gravity is 9.81 Multiplied with hell taxes. Even a 0.6 m multiplier to be prepared by sign. Thank you. Pretty angry. So from here we get bogged down by gravity as equals 2 -1.9 Julie. This is the work done by the gravity now in party. And partly we need to calculate the net bogdan on the block so that the network done will be equals two. W net will be equals two. Work done by the force was bogged down by the gravity, but it will come out to be 1.9 -1.9 that is equals to zero jule.


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