a-2000-kg-is-initially-traveling-with-a-speed-of-20-ms-the-driver-applies-the-break-and-the-truck-sl

Question

A 2000 kg is initially traveling with a speed of 20 m/s. The driver applies the break and the truck slows to 10 m/s. How much work was done by the frictional force applied from the breaks?
(A) –10,000 J
(B) –5,000 J
(C) 5,000 J
(D) 10,000 J

Vishal Gupta · Accepted Answer

Hi in the given problem mass of the object is M is equal to 2000 kilogram. It&#x27;s initially speed is given to be v. I. is equal to 20 m/s, and the final speed is reduced to do by applying bricks that is reduced to and meter per second. Now we have to find work done by the frictional force. So using work energy theorem that work done by the frictional force will be given by change in kinetic energy knees. Final kinetic energy minus initial kinetic energy means this is half and we every square minus half and the eyes square. So plugging in or loan values this we have done is equal to we can take this half M. As a common out for em. This is 2000 kg for we have this is 10 m per second square minus 20 m per second. The square. So it comes out to be 1000 bracket 100 -400. So this is finally 1000 multiplied by minus 300 Jews. So we can say this organism minus which can be represented as minus three into 10, part five jews minus three luxury, or minus 3 to 10 days. Part five jews, which is the answer for this problem, but none of the option is correct here in this problem. Thank you.

Manish Kumar · Answer

wait off the card is given and that it tickles too. The mass of the car, which is equals to 12.500 to the power of three kg And what I&#x27;ve done on eighties given with his 5 kg which is 5002 jewels. So basically, we have to come from the final speed in the first part. So we know that by the bar garage term work done Well, way change. It can technology that will be in a Socratic in the zero. So the federal energy we have half of the square. So what we&#x27;ve done is five 1000 is equal to one upon two into 2.5 into general, two out of three. So this and to be a square. So this one and this one get canceled and multiple ever too. So there&#x27;s really on five. So this basically this one is going to get can seal. So be a square. He become nothing but 2 m parts again. And in the next part we have to get a great force on the on the excited on the car. So basically, the distance this one is given distance traveled in this time is even That is 25 m. So we know that b squared. You will do nothing. What you square plus to S and B zero in the U. S. Zero in this case would be a square you&#x27;ve got great is visit for that is about two times off a exploration into distance 25 this is and we can say that the acceleration comes out over there. This is was poor upon 15 m person going to square. So the force we have to learn in this will do nothing but mass times acceleration. That could be that will be 2.520 year off. Three times off, poured upon 50. So we can say this will become too over there. This will too. So we can say that to inter Tendons were off to Newton Force exerted on the car

Vishal Gupta · Answer

hi. In the given problem, mask off the car is given us 1000 kg here. This is the horizontal surface over which this car is moving with an initial speed off really is equal to 5 m for second. And after applying the brake, this car has to come to stop. Finally, so it&#x27;s finally speed is zero and it has to move our distance off. 20 m. A car moves a distance of 20 m before coming to rest. If it&#x27;s smart, is M, it&#x27;s weight is MG. Acting vertically don&#x27;t work. So the force of friction acting between their tires and the ground is given us new times off its weight. Their new is the coefficient of friction and using Newton&#x27;s second law of motion, this force acting on the car in this opposite direction, opposite to the direction of motion. He&#x27;s given us em into a so canceling this m the retardation produced in the car comes out Toby new into G or we can say this is 10 mute and this is the tradition. So we associate a negative sign also with it now using third equation off motion, which this is we f square is equal to V I square plus two a s. We have this zero v I. This is squared off five for a This is minus two into 10 into Mu s is 20. So finally it becomes 400. Milk is equal to 25. So immune comes out to be 25 by 400. Or we can save coefficient of friction between the tires and the ground is 0.6 to 5. Means your option de is correct. Hope it. Answer your question. Well, thank you.

Ze-Han Lee · Answer

So in this problem, we have a car. And after we hold the breaker Ah, After we hold the break, we say that Ah, uh, it keeps going to see escapes forced instant distance off 20 meters. So, uh, the mass of the car is the 1000 kilograms and initial speed of the car is flying meters per second. So, uh, and oh, the skate biscuit. This is just ah, 20 meters, so we can list the equation. Ah, which So the inner stroke magnetic energy of this car equal the work done by the friction. Right? So this becomes g times mu k times x, some jihadism uk is the friction and X is determined to be Ah huh And the squid be Bye bye. Sorry. Not eggs. Okay. They were by m g time. Sex. Okay, so this gives me a point. 06 to 5. All right. No,

Suzanne W. · Answer

Okay, so we have 1000 killers, kilogram car pushes a 2000 kg truck that has a dead battery. I&#x27;m a driver steps on the accelerator. The drive wheels of the car push against the ground with the force of 4500 Newtons. Rolling friction can be neglected needed. The magnitude of the force of the car on the truck and the force of the truck on the car. Okay, so let&#x27;s look at a free body diagram. So you have the car, it has a normal force, It has the force that is pushing on C. You have another force, so the surface onto the car. You have the force of the truck on the car. Then we also have the force of gravity, and with the truck also have the normal force the force of the car on the truck, and you have the gravitational force so notice there&#x27;s no surface on this one. Your action reaction pair is truck on car to car on track, which is what we&#x27;re looking for. So where we are summing everything in the X. Direction. So the net force in the X. Direction is F. S. On C minus F. T. On C. Which is the mass of the car, the acceleration of the car. And for the truck it is the force C. T. And closed the mass of the truck, the acceleration of the truck. So knowing all of these, we know that s the force of the truck on the car is going to be equal to the force of the car on the trucks. And so action reaction pairs. So we can say that the force of the car on the track is or over the mass of the truck is equal to the acceleration because the acceleration of the car is equal to the acceleration of the truck. It&#x27;s just the acceleration. So plugging this in, we know that combining this equation and the fact that FC on us minus, let&#x27;s see on teeth divided by the mass of the car, is also equal to the acceleration. So this gives us that the force of the car on the trip is equal to the force of the car on the surface finds mass of the truck over massive a car plus mass of the check. So, plugging in, we know that this is 4500 newtons times 2000 kilograms over 1000 plus 2000, which is 3000 kg, which is three 1000 students. And this gives us our answer. & dude, in Newton&#x27;s 3rd Law, The force of the truck on the car is also equal to 3000 Newtons.

Paul A. · Answer

Okay, so we have a model car and it&#x27;s controlled by a radio, and it&#x27;s moving in the ex direction. The motion is directed according to the Force Displacement Rough Tzar FX, and this is N uh X is M. And so the graph is divided into specific units one two on negative one negative to that tells us that the force faces in different directions the motion off the cars modeled by this specific diagrams. So there&#x27;s a triangle right there and then a trap his own between a displacement off zero and three, we have a triangle and then three, actually zero two between zero and two. We have a triangle. This is right here and then this is three because all the way up until four goes in the negative direction, that&#x27;s the force in the negative negative force and then goes all the way up until so this point is a displacement off six. And then right here, the displacement off. Seven. So was supposed to find the velocity off the car when X equals to zero point three meters when X equals two four point or rather not zero point three mirrors, but three point zero meters Exit polls to four meters and then X equals two, uh, seven point zero meters. So we&#x27;re using the work energy serum R. The walk is equivalent to the change in Connecticut. Aji so w it calls to one half empty final squared minus one half. Have we initial squared? Uh, we&#x27;re told that the car starts from rest when X equals to zero. So that means that this part of the problem will have an initial velocity off zero. So v I equals to zero meters per second. And that means that the final kinetic energy will be zero. We don&#x27;t have to account for that. We just need to find the walk down in moving the the car from the point zero all the way up until the point three. And that&#x27;s the area off the proper side. Because, remember, work is force times distance. The Y axis is the force. The X axis is distance. So we&#x27;re going to do one half time&#x27;s bays, which is from zero to two. So two meters and then times the height, which is to Newton&#x27;s. That&#x27;s the energy expended from zero to two, or the energy required to move the car from zero to two. And then we also have the rectangle. This one this part which is just the base times the hay. The base happens to be one meter from two to three, and then the height is too, Newtons. And so the total area these to cancel out, we have to Newton Meter, um, plus to Newton meter. Remember, New comedian is another name for jewels. So we have four Jules, then one house M V squared every final squared. We want to find the final velocity. So that&#x27;s the final squared equals to the fruit. Divide both sides by one half em. And this one is one. Have him remember the mass of the object happens to be two kilograms s. So we wanna plug in those numbers there. This two comes up and changes that to eight. So we have eight jewels off, uh, two kilograms. That&#x27;s thievy Final squared, and it&#x27;s going to give us a velocity. You know, it&#x27;s going to give us a final velocity eight over, too. That&#x27;s full. To find the the final. We have to take this quote of that So eight Jules off to kilograms and that gives us of the final velocity off squirrel off H. Now we&#x27;re just going to do that with the car potatoes. This quote eight divide by two and that gives us two. So two meters per second. Okay, gives us two meters per second. So once again, let&#x27;s just follow the steps. The energy energy, we&#x27;re required to move the car from zero to No, from zero meters of the origin to three meters is for Jules, and we have we have the the final kinetic energy which is one half empty, square, empty, final squared. And so that&#x27;s going to give us a V finals quote of a Jules off two kilograms of the second part of the problem is they want us to find the velocity and I&#x27;m going to do that in the next paid. So we want to find the velocity when X equals two four point zero meters. Members could look at the diagram, you notice it goes up and then goes down and then, um stops at four. So there is no change in energy. So we have zero jewels between three and four. So that means we still keep the same four Jules from before and, um, for Jules and then one half and v squared were using the work. Energy, Kerem would say, is what it was to change in kinetic energy. So what it was to one half M V. Final squared, minus one half M. V initial squared the initial Connecticut zero since the initial velocity it started from Resta. Initial velocity happened to be zero meters per second vi final squared, one half empty, final squared. We divide both sides by one half em and one half em. So it&#x27;s similar to the previous problems. So we have the final square equals two. These two comes up and don&#x27;t forget the mass, which is two kilograms, the muscle, the car model. And so we have a Jules off two kilograms. We get the square root of that V Finals quote off a Jules over two kilograms, and that gives us the final velocity off two meters per second again. And that&#x27;s the velocity art. Four meters. So this is X, and this is our force in Newton&#x27;s. So the velocity at formulas also happens to be two meters per second. And this makes sense because between three and four. We don&#x27;t see any additional energy. So we still have the same old energy off for jewels that happened between zero and three meters. So we&#x27;re using that and the object still has a mass off two kilograms doesn&#x27;t change. The mass doesn&#x27;t change the last part of the problem. We want to find first of all, the energy from zero meters, two, seven meters and then use that to find the velocity. You know, what&#x27;s the final velocity with X equals two seven point zero meters. So this is point zero. We go backto a bagman. It goes up and then down up until full. So this is from zero to two and then two to three and then three to four. And then between foreign six, it changes toe rectangle and then its flight all the way up until seven. This is the exact CE is the motion of the orb of the car. And this is the Y axis. The force involved so fast. Of all we have to we&#x27;re using the work energy theorem says that walk is the same was changing kinetic energy. So the same walk is one half m v final squared minus one half and the initial squared. Our life is a little bit easier because the initial velocity was zero. So that zero meters per second and so all we have to do is find the work and we already know that&#x27;s the energy between zero and four happens to be for jewels, because this part was for jewels from the previous. From the initial part, you can see going back when we did the computations, we found out that the energy involved is all of that, you know, to Newton meters plus two new meters and then in the second part&#x27;s the energy involved between three and four happens to be zero jewels because we don&#x27;t have any changes in the force. The force itself affects zero. And remember, the energy is forced Time zero this times distance, even if you&#x27;re changing distance and your force zero mutants, you&#x27;re gonna end up with on energy off zero jewels. So that&#x27;s the second part. The only thing we need to compute is the energy from full to seven s o going to do the energy from four to six. It&#x27;s a triangle. So one half base is forty six. That&#x27;s two majors and then the height is negative. One Newton&#x27;s. So the energy becomes this to cancel out. So negative one. Jules the energy from six to seven or the work done in moving the object from six to seven happens to be zero Jules, because we don&#x27;t have any force acting off. There. You see on the Y axis the force of this point zero mutants, and if you want to find the work, you always do force times distance. Zero Newton&#x27;s times uh, one meter is just going to give us sirah. Jules. So this part gives a sigil sew were summing up all the energy, the total energy Oh, the total work done from the beginning. So between zero and three, it&#x27;s for jewels, plus between three and four meters. It&#x27;s zero jewels and then plus between foreign six meters is the area of this triangle. So that&#x27;s negative. One. Jules. And then between six and seven. That&#x27;s right here. It&#x27;s zero jewel. So the total walk down we just combine the foreign, the one Jule and we get three jewels. We&#x27;re going to use that in the walk energy yarn. So we have three jewels. He calls to one half. Then the final squared the divide both sides by one half em. And this is also one half em VI. Final squared becomes the same as this to comes up algebra. So two times three is six tools over the mass, which is two kilograms. So the final equals to this quote off. Six. Jules over two kilogram, six over to his three and using my cart later, we&#x27;re going to take the radical, so this is screwed off. Six divided by two. And that gives us one point seven to read, So the velocity at X equals two seven is one point seven three meters per second. So just a quick Rick up off the process and finding the velocities at specific points, the first thing you have to do is to remember that you find the energy the walk down in moving the object from zero to three and then from three to four initial, it&#x27;s from zero to three on. There&#x27;s a triangle on the rectangle, so finding the area&#x27;s off, those two is going to give you the work done in moving from zero to three. Because Thie y axis is the force. And the X axis is thie our distance, by definition, walk his force times distance. And we got that energy, Toby for Jules. Now compare that to the work energy serum, the initial energy, the initial Connecticut He happens to be one half envy. Initial off square, the initial velocity zero. So we are counting. We&#x27;re cancelling that out. This is going to be zero energy. S O B final, as you can see, is justice. Quote off. A Jules off two kilograms on that gives us two meters per second squared in the second part of the final velocity is going to remain the same because we don&#x27;t have energy. Any walk expended or ex expanded between three and four, it&#x27;s zero jewels. So we have the same velocity. Nothing changes because there&#x27;s no work in going on. And then in the last part of the problem, we have to find the energy at each section off the displacement. So displacement from zero to on three years for jewels from three to four. Residuals from four to six is negative one, Jules, because the force is going in the opposite direction. And then between six and seven, it&#x27;s originals. We do the math. The walk energy term Counsel&#x27;s out. The initial kinetic energy, which is going to be zero on DSO only left the final Connecticut and you use that. Compare that to the total war we got. Total walk was three jewels. And so after the math, we get a final velocity at seven off one point seventy metres per second. Hope you enjoy the video. Feel free to share any questions and any questions out and looking forward to the next problem and enjoy a day. Okay, Thanks. Bye.

Problem

A box of mass $m$ slides down a frictionless incl…

Problem 15 Easy Difficulty

A 2000 kg is initially traveling with a speed of 20 m/s. The driver applies the break and the truck slows to 10 m/s. How much work was done by the frictional force applied from the breaks?
(A) –10,000 J
(B) –5,000 J
(C) 5,000 J
(D) 10,000 J

Answer

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