A man Poos a crate of mass 73.0 69 2808S5N across a level He 28815 with a force at an angle of 0o above the horizontal_ When the crate is moving, the frictional for...

A 36.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m the floor is frictionless, and for the next 10.0 m the coefficient of friction is 0.20. What is the final speed of the crate after being pulled these 21.0 m?

Oh, for party. We can first say that the sum of forces in the ex direction would be equaling the force applied minus the force of friction. In this case, we can say that here the force applied. We know the force applied for the force of friction. We can say minus um you the coefficient of kinetic friction multiplied by the normal force. In this case, we know the normal forces equaling mg. So this would be the force applied minus mu M g. And in this case, we considered then the net force in the X direction which would be equaling the force applied 275 or Newton's most rather minus 0.358 multiplied by the mass of 92.0 kilograms, multiplied by 9.8 meters per second squared. And we find that this is equaling negative 47 0.77 Newton's And so this would be the net force on the on the crate. While the crate is in contact with that rough surface, that would be the answer for part A. Now, for Part B, we want to find the net force the work that the Net force does on the crate. And so we say that the net, the work done by the net force, would be equaling the magnitude of the net force in the ex direction multiplied by the displacement in the ex direction or the length of the table. And so this is equaling negative. 47.77 Newtons multiplied by 0.65 meters. And this is equaling negative 31 point one, Jules. And so this would be the, uh, work done on by the net force on the crate for parts be for port. See, Now we want to find the final velocity after it reaches the end of the rough surface. And so we know that the net force in the X direction is equaling the mass times acceleration in the extraction. So the acceleration in the extraction would be equal in the net force in the extraction divided by the mass of 92 kilograms. So this is equaling negative 47.77 Newton's. This would be divided by 92.0 kilograms and we find that the acceleration in the extraction is equaling negative 0.52 meters per second squared and so we can say that velocity final squared equals velocity Initial squared plus two times the acceleration times Delta X and so the final velocity is gonna be equal in the square root of the initial velocity 0.85 meters her second quantity squared plus two times negative 0.52 meters per second squared multiplied by delta delta X of point 65 meters. Stand the square root and we find that the final velocity is gonna be equaling point 20 meters per second. So at the end of the rough surface, the velocity of the crate has decreased from 200.85 meters per 2nd 2.20 meters per second. This would be the final velocity of the crate after it reaches the end of the rough surface. That is the end of the solution. Thank you for watching

So here the work total is going to be equal to the work. Done. Initially worked unsub to work down. So what dunce of one plus worked on some two minus the work done to friction. So this will be equal to the force. The pulling force Times dese of one co sign of zero degrees. This is, of course, equal to one plus the pulling force. Times decent too. Co sign of sir degrees. Again, this equals one. And then this will be plus the force of friction Another plus the force of friction decent to co sign of 180 degrees. So this is where it becomes negative. And so we know that the work total is going to be equal to the change in kinetic energy as well. And this legal 1/2 Mm, the final squared minus the initial squared. And you know the initial velocity zero. So we can say that the final velocity is going to be equal to two times the pulling force. Time's dese of one plus de so to minus the coefficient of kinetic friction times the mass times the gravity times decent too. And then this will be divided by the mass all to the 1/2 power and at this point, we can solve. So the final velocity will be equal to two times. 225 Nunes times 21 meters minus 210.2 times 46 kilograms times 9.8 meters per second, squared times 10 meters and then this will be divided by 46 kilograms and then again quantity to the 1/2 power and the final philosophy is found to be approximately 13 meters per second. So this is our final answer. That is the end of the solution. Thank you for watching.

For this problem on the topic off work and kinetic energy, we're told that it creates. It's at rest at the bottom, often inclined plane before a force is applied to push the they create up the ramp. The the ramp exerts a frictional force only create, and we are asked to calculate the work done and the create during its motion, from the bottom to the top of the ramp, as well as the time it would take for the great to travel to the top of the ramp. So let's firstly look at the situation. Now. We know therefore forces acting on the crate, the force due to the push, gravity friction and the normal force due to the surface of the ramp. So let's first calculate the work done by each of these four forces. Now the normal force does no work, since it is perpendicular to the displacement off the crate, but each other force, we'll do work. So the work done by the push we call it W. F is equal to the force. Applied F times, the co sign off 34 degrees times the distance up the ramp off 15 meters, and so This is 290 times the call sign off 34 degrees times the distance of 15 m. This gives us the work done due to the pushing force to be 2606 jewels up the ramp. Next is the work done due to the weight of the create W M G. So this is the weight minus M g, which is the force times the distance or 15 meters times the component perpendicular to the slope. Sign off 34 degrees. Thanks. So now we can full in our values. This is minus in. The mess is 20 kg will suppress the units times G, 9.8 m per square second and the distance off 15 m multiplied by design off 34 degrees. So solving we get to work done by the weight his minus senses in the opposite direction to the push minus 1644. Jules And finally we worked on YouTube fiction acting against the motion off the create is minus the frictional force. Little f claims the distance 15 m. So again we have this value so we can put this in this is minus 65 Newtons and the distance off 15 m, which gives us the work done due to friction to be minus 975. Jules. So now that we have the work done by each of the forces, we can calculate the total work as the sum of each of these values. So the total work, then w T is 3600 and six jewels minus 1000 644 jewels minus 975 jewels, which gives us the total walk down in this process to be 987 Jules. And that's our answer for part A. We calculated the total What time for this process? Now we want to know the time it would take for the creator travel to the top of the ramp. So we first need to find the final velocity. And we know that the total work ban on the crate is most equal to be change in kinetic energy that a k which is ah half am VF minus. We have squared minus V I square. So since the initial velocity is zero, we can rearrange this equation and solve for the final velocity V f. And this is simply the square root off two times the total work, which is to times 900 and 87 jewels divided by the mess off the create, which is 20 k g. So this gives us the final velocity top of the ram to meet nine 0.9 35 m per second. Now we can use this since we have a constant acceleration in the equation, you kind of medical equation X minus X note must equal to V not ex, plus the final velocity in the X direction VX divided by two times Time t so rearranging we can find the time off. Traveled to the top of the ram to be t is able to into X minus. It's not over the not ex plus the X, and since we know the final velocity, we can calculate this time. So that's two times the distance off the slope off 15 meters off. The initial velocity we know is zero. That's the final lost. You calculated to be nine 0.935 meters the second and so calculating. We get the time it takes for the crate to move from the bottom to the top off the ramp to be 3.2 seconds

Okay, so this is Steve Tire Gradual. For this question, As you can tell, the crates is rotating about point eat. And since the crate has ah shape of a rectangle So we know that we're here is the with off the wreck tango. And now here is the length off the rack tango, and say the one here is equal 25 degree. Okay. Rt here is the, uh, cover and equal distance. Call attention to a point he wishes the Dev are of attention. RG here is the per vertical distance for gravity to appoint me which is the deliver of the gravity. And when attention is having a mango, um, about horizontal surface, we just which is about 61 degree. So therefore we'll have there too. Is he going to attention to about anyone now? We're seeing signs w over to over over two. Okay, which is equal to tension. Uh, negative one that we all well m and A w which is the with off the, uh, rock tango. A stroke win four meter in the land off the rack. Tango is 0.9 meter. Which alley? What was your point? Nine year. So if we plan came back to the question where angle fair to it's eventually going 24 degree. So therefore, with RG over the high pasta news off the triangle, O E is equal to concentrate at one place. They are two and left off the high positives off Try and go. You seek to ah, square root that we over two square plus hours who square So therefore, how RG was this? The pumpernickel distance off the gravity to the point is equal to co signed they that wont plus they doubt to times square root w over two square over two square And we know the value for Taylor won the world for state out to and running for the w an hour So you mentioned Have RG is equal to consign 25 degree plus 24 degree Times Square You 2.4 meter over 202 plus uh, 0.9 meter over to power to and this will give us Archie, you see gold 0.32 meter And when our c o l can we go to side beta Okay And Breda is you go to 61 degree minds minus fatal. Okay, so that is here. So therefore, we have rt zero al sighing 61 mine and stayed on, so we know they always Tony, my degree I was your point. I meter. So there will rt is he will drop 610.0.53 meter and they the torque that was produced by gravity should be do the fourth that was produced by attention. Okay, I was Really Why is that? Well, because, well, considered eso, as you can tell the crazies rotating about pointy. So there, for the moment about money, which is some Asian and me here each other moments off the they told him was about point. Should we go zero okay. And this can be go to, um, TRT minus G R g. Okay, because we'll consider the moments that is in a counterclockwise direction is positive. And the moments in the car cars directions is negative. So, as you can tell, let's go back to the graph. So, as you can tell, gravity is rotating in a clockwise directions. Whence about pointy. Okay, and attentions is rotating in a calm, clockwise direction. About 20. So that's that's why the moments now what a torque produced by the attention is positive and the torque produced by the grab these negative okay, So eventually will have g r g look is equal to TRT, which is the equation here. So they're of of attention T here is your g r g o r t and we know G, which is a gravity can be expressed as mg So have attention is equal to Emma Grz over rt So we know the mass over a crazy 72 kilograms and the acceleration of gravity is 9.8 meters per seconds Were RG 0.32 meter rt still 0.53 meter So therefore the attention can be able to, um 72 kill worms turns nine point a meter per seconds were joe 0.32 meter over Jo Pilyeon 53 meter and it will give us the attention is because who? 426 Newton. So if we run it off, we right off to the 10th digit we'll have detentions is approximately equal to 403 430 Newton. Okay, and this is the answer for this question.