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Ch 07: Newton's Third Law

Chapter 7, Problem 6

A 50,000 kg locomotive is traveling at 10 m/s when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop? Assume the track is level.

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Hi everyone in this practice problem, we will have a 350 kg rubber barrel rolled on level concrete ground at 5 m per second. We're being asked to determine the distance the barrel will cover before it comes to a stop or before it halts. And the options given are a 63.8 m b 67.3 m C 83.6 m and D 76.3 m. So we will use the particle model and consider the barrel as the particle. Let's first look into the pictorial representation of the given situation and I am going to just indicate the barrel as this circle right here. So the first uh force that is going to be acting on the barrel is the gravitational force or FG, which is just essentially the mass of the barrel multiplied by the gravitational acceleration. And then the countering that we will have the normal force. And here and then the other force that is going to be acting is the frictional force, which is going to be represented by this arrow right here on the left, which is just FK right. So just like so awesome. So the barrel will have a rolling friction. So we will use the coefficient of rolling friction for rubber on dry concrete from the table that we have, which is just essentially mu K oops mu K is going to be equal to 0.02. And that is obtained from just the coefficient of friction for rubber on dry concrete. Awesome. So the barrel is going to be moving in the horizontal direction. Hence, from the free body diagram that we have here, we will get Sigma Fy to then be equal to N minus FG which will be equals to N minus M multiplied by G which will be equals to zero because it is not going to be moving in the uh vertical direction. So uh rearranging this N will be equal to M multiplied by G. And then looking into Sigma FX, we have Sigma FX to be equals to M multiplied by A X which in this case, Sigma FX is just going to be FK and FK is going to be equal to UK multiplied by N. If we recall the equation for friction, uh friction force and UK multiplied by N will be equal to M multiplied by A X. So we can actually substitute our information that we know UK is 0.02 N is M multiplied by G. So I'm just gonna keep it as that for now M multiplied by A X is M multiplied by A X and we can cross out the MS. So therefore, we will get an equation where A X will be equal to 0.02 multiplied by 9. m per second squared. Just like. So and then calculating this, we will then obtain the acceleration to then be equal to 0.196 m per second squared just like so awesome. So we know that this acceleration right here is essentially what is causing the barrel to stop rolling. So the barrel is going to be going in the velocity heading to, for example, the right side here, this is the velocity and the acceleration is what is going to be stopping it from rolling, which is A X. So in this case, then we know that A X is going to be negative. So I will put the negative sign there negative 0.196 m per second squared. So now that we know that we want to apply kinematic equation which will then be equal to VF squared equals to V I squared plus to A X delta X. And we want to note that it will come to a full stop. So VF square or VF will then be equals to zero. And then I'm just gonna cross that out now equals to zero. And then the delta X is what is being uh calculated or unknown. And the A X is known in the problem statement V I is given to be 5 m per second. So we want to substitute that into our equation. Awesome. So let's start doing exactly that. So we know that the VF squared is going to be zero. So I'm just gonna write down 00 will be equal to V I squared, which is going to be 5 m per second squared just like so and then plus two multiplied by A X and A X we have obtained previously which is negative 0.196 m per second squared multiplied that with delta X delta X will then be equal to. So 5 m per second squared is going to be negative because we're gonna put that into the other side. So negative 25 m squared divided by second squared divided by two multiplied by in parentheses, negative 0.196 m per second squared. In calculating this, we will get or obtain delta X which will come out to a value of 63.8 m. So that will be the final uh answer that we will want for this practice problem. So the total distance that the barrel will cover before it comes to a full halt is going to be 63.8 m which will be corresponding with option A. So option A will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion. Please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this video. Thank you so much for watching. Bye bye.