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What is the net force on a 500 gram cart which experiences two forces, 0.35 N to the left and 0.14 N to the right?

b) What is the acceleration of this cart?

Hi everyone here it is given a card. Uh huh. Yeah. Yeah. Yeah. Yeah right. Of must 100 cases filled with the whole yeah 1800 cages. So total mask baby that is 1900 Katie. It is pulled by the force 14 colonia to or 1400 14,000 newton. Up along and cracked. Played having inclination 23 11 and your cement is given. Okay it won't find me dead. We have to calculate Brooklyn by armed forces. Uh huh. Foreign courses and velocity at the top. Yeah. Forces you can draw. It's a bit MG vertically downward normal component of it. I. M. D. Cost tutor. Am you sent it to? Yeah. And fiction. Normal force will be empty. Costed fiction. Maybe you into N. That is immune. MD. Costed now you can calculate the workmen but by the first case worked in by applied force. Have a into the cause of the room force is 14,000 and two distances. 10.5 m into one. So it is to be 14.7 10 to the power five Children. Work done by gravity. Mhm And GT because of competing masses 1900 G. Is 9.8 cause of 23 and took D. D. Is 10.5. So you will let 1.836 10 to the power fight judy government by friction minus sorry mule and D'Costa. The cause of quality muse for three MG is 1900 Gs 900 sorry. Cause of 23 2 and when fight. So it is to be minus. Yeah. Uh huh. 0.55 10 20 About five to yeah. Government by MG costs it up. And he said sorry welcome bye norman and in two people because of 90. So it is zero period. Welcome bye and design theater. I am design theater in two D course of vanity from minus m 1909th. Partied sign of 23 into deep and want to fight. So just do you 0.779 10 2 people five. No. Now we have to find the velocities of using dark energy terra net. Golden. It's got to gain in kinetic energy because it is starting from rest. So never worked in between 14.7. Okay. 10 to the power five minus 155 10 to the Power five minus point 779 10 24 5 of em 1900 into the square. So velocity you will get 26.2 people mr per second. Hey, there's one. Thanks for watching it.

In this problem, a person pushes a 16 Katie shopping cart at a constant velocity for a distance of 16 m. She pushes a direction 25° below the horizontal. At 42. Newton fractional horse opposes the motion of bye. Okay, so it is us. That what is the magnitude of the force this shop? The person shopper exert straight the shopper exit on the shopping cart. Right? So we need to determine the work done by the pushing fools, right? And they can be disaster work done by frictional force. And we are asked to find it. Work turned by reputational force. Yes, So this is all about angles making with pushing force, frictional force, and gravitational force. Right? So we are just calculating for it. So here on applying Newton's second law of motion, we know that Net Force is a mission of f equals two. Of course 25 minus if he goes to zero. Right? So this will give us upon F Small left by past 20 five. So we're small air is what? Small service. 42, New Daniel. Okay, so this will give us 42 divided by two. Was 25. This will hear us What is six Yeah, today forward, you know, So this is it answered. Okay, now, for the second part, work done by pushing force will be equals two. Okay, W one FD W one equals two. FD cost 25. So 42 into 16 cost 25 with the word. So he and legacy for 25. Engine 16 and 3016. There to play 16 to what you do. Engine boss 25 great. So this is going to be risk 609 16. 609 0.3 plane then this is this is all Jr together. Yes. So they should win and you expect this is it. And right for the C. Part work done by the by the frictional forces, work done by the frictional force will be it was too W It was two FD because one intuitively, you know that the angle between them is one degree. We know that if it is an object then if if it is moving in this within frictional forces acting this week with the angle will be went into Disney like. Mhm. So here we can see We can say 42 into 16. This will give us minus 600 72. I'm sorry, I need to correct here somewhere. Uh this is absolute pushing forces. You see the the angle is zero taking the year. So this will give us mhm 4240 is 1 42- 16 years. Six 70 two. Yeah, late. And We are getting by 6 72. Were done by original courses. So, these forces are pushing and pulling forces. Are You put an opposing the artist pushing one is 673 two and French forces -670 two. Right, so now last for the last one. Love done by gravity will be equal to F. D. Cost like right. So they should be Indeed, they should be MD by engine 20. That is zero, joe. It's great. So, this is an answer. I hope you understand. This is how we solve this problem. Thank you for watching.

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Question will help you in a car which is of Masi. Um We have given the masters 500 g. If you can work this into Casey, this is nothing but half Casey Mears point for kids. Now this cart is acted by two forces, One forces acting right side with the magnitude of point 1/4. Newton. And the one is acting towards left side with a magnitude of 10.35 Norton. So now we have to find the net force definitely call the how much And what is the acceleration with which the cart most? No, yeah, definitely is nothing but some of all forces along that direction. So if you take any direction then some of all forces Gusti net force along that direction. Yeah. No. Since the forces are acting in only one direction, that is only in horizontal direction. We have only one F net in horizontal direction. So in this case f not equal to ah left hand side force, left hand force minus. Sorry. Place a right hand force. No, we have to specify that we have to take any one direction suppose. Consider let us consider right side direction. That's positive means the force acting the forces acting along this direct not pashto and forces acting along this direction, not negative. No left hand forces acting towards left. So it is negative 135. No time right hand forces acting towards right. So it is poised to at this 0.14 note on. And we get it has minus point 21 note these 21 note. And so these negatives and indicates that the forces acting towards left. These negatives and indicates the forces acting towards left. So we can write this point to a Norton towards left. No, I was trying to find the acceleration of this system as you all know that from the Noten's firstly Norton's first lor the networks acting in a body equal to mass times, acceleration where m is the mass and the A. S. D. Acceleration. Yes, So we got definite as .21 and Mars has .5 kids is how to substitute Always mass in terms of cases only enforcing Norton's only Now we get equal 2.210.5 If you do this it get .42 meters per second squared. This is the unit of acceleration. So uh no we have a doubt in water direction. This acceleration is okay since it is paused to do you think this is right said no, the acceleration always act, acceleration is always in in the direction of net force. We have net so what is the direction of definite here regard? It is left said. So the acceleration we got here also towards left side. So we finally we the cut is moving With an acceleration .42 m/s squared towards left side