A 2.0-kg block slides across a rough surface with a constant c...

Question

A 2.0-kg block slides across a rough surface with a constant coefficient of kinetic friction of 0.50 (Fig. 7–38a). The block starts at x= 0 with an initial velocity of 4.9 m/s. Pushing the block is a force directed at 36.8° below the horizontal and whose magnitude increases with position as shown in Fig. 7–38b.
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(d) Draw a line on the graph showing the magnitude of the friction force versus distance x.

(d) Draw a line on the graph showing the magnitude of the friction force versus distance x.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A metal sheet with a mass of 3.0 kg is being pushed over the floor of a factory figure one. The floor offers a steady resistance to motion known as kinetic friction with a coefficient of 0.35. At the beginning, the sheet is at the position marked as X equals 0.0. And it's moving at a speed of 2.0 m per second. The force used to push the sheet is angled 15 degrees below the horizontal and this force grows stronger. The further the sheet moves, which is illustrated in figure two, draw a line on the graph showing the magnitude of the friction force versus distance X. So that's our end goal. Ultimately, we're gonna be drawing a line on the graph that that shows the magnitude of the friction force versus the distance X. And this is denoted by figure three. So we're gonna be using this skeleton graph right. Now because it doesn't have a line yet because this skeleton graph that's blank in figure three, we're gonna be using figure three to draw a line to find our final answer. Awesome. So before we dive in and start solving for this, let's examine our figures one and two really quick here. So figure one is literally a visualization of what we just discussed in the problem itself. So we have this sheet metal, which is denoted as or AAA metal sheet, I should say, which is represented by this green rectangle that's up on top of this other rectangle, this white rectangle which represents the floor. So we have this metal sheet that's 3.0 kg and it is being there's a force that is being used to push this sheet and it's angled at 15 degrees below the horizontal as shown because this dotted line denotes the horizontal. And the blue arrow shows that it's 15 degrees below the horizontal and it's being pushed at a speed of 2.0 m per second because it's moving at 2.0 m per second and it starts from rest at 0.0 m and it will be moved 2.0 m ultimately. And we could see in figure two, which you could see that the force being put using. So the force that's being used to push the sheet at 15 degrees below the horizontal and this force will grow stronger the further the sheet moves which you could see. And this linear graph it's just slowly increasing as the, as the distance gets further away. So from 0 to 1.0 to 2.0 the force will just get even stronger. OK. Now that we have a visualization of what's going on here and then looking at quickly at figure three here, we have our, our friction kinetic friction of in Newton. So like FK of N is our vertical axis and then our horizontal axis is X of M which is similar to figure two where our vertical axis is F of N and our horizontal axis is X of M. OK. So now we have an idea of what's going on here. We need to make a couple assumptions first to solve. So firstly, in order to solve this problem, we need to make a couple assumptions. So our first assumption, we need to assume that the coefficient of kinetic friction will remain constant throughout the motion. And secondly, we need to assume that the surface of the factory floor is level. So when we calculate the sliding friction value, we will only consider the normal force. So now let us create a free body diagram to help us better visualize all the forces at work in this particular problem. So looking at our free body diagram here, we could see this time things have kind of slightly changed here. So in this case, our sheet metal is denoted by this peach colored rectangle and then the gray rectangle will be the floor. So going straight up with this blue arrow is denoting our normal force. And then we have a peachy orange arrow pointing straight down, denoting our weight, which weight is the mass of the sheet metal multiplied by the acceleration due to gravity. And then going to the right, we have a red arrow which is denoting our force that's being applied to the sheet metal. And of course, in the opposite direction of the motion, we have our friction, our kinetic friction force which is fighting this motion of the sliding metal sheet across the floor. And in our vertical direction is the Y direction. And then in our horizontal direction is the X direction. And we have in the X direction, we have our F of X value was denoted by a pale green arrow. And then our other pale green arrow going down is our F of Y force going downwards in the vertical direction. So going to the right on the horizontal plane is our F of X value. Awesome. So now that we have a visualization of the free body diagram of all our forces that are at work here, we need to note that the kinetic friction is not constant under these conditions because the normal force N is also changing by a magnitude of 10 X as depicted in figure two. So we need to note that we will plot our value over the specified range X equals 0.0 m to X equals 2.0 m. So now we need to recall and use Newton's second law in the vertical direction to write the following equation that N minus W minus FY is equal to zero and rearranging the sulfur N all by itself, we get that N is equal to W plus F of Y. So now we could plug in all of our known values. So we know that N is equal to, so we know that the weight is equal to the mass multiplied by the acceleration due to gravity plus 10 X multiplied by sine of 15 degrees, which is the angle that's provided to us in the prom itself. OK. So now at this point, we need to recall and use the kinetic friction equation which states that FK is equal to mu K, which is the coefficient of kinetic friction multiplied by N which is equal to mu K multiplied by M multiplied by G plus 10 X multiplied by sign of 15 degrees. Awesome. So at this point, we could plug in all of our known values. And when we do that, we will get that F of K is equal to. So F of K, so the friction kinetic friction is equal to 0.35 because that's our coefficient of kinetic friction multiplied by 3.0 kg because that's the weight of our me or I should say the mass be specific, the mass of our metal sheet multiplied by the acceleration due to gravity which is 9.8 m per second squared plus 10 newtons per meter. X multiplied by sign of 15 degrees. OK. So now we need to take this equation. So to avoid confusion, let's write this down one more time because this is very important. So our kinetic friction is equal to 0.35 which is our coefficient kinetic friction multiplied by the mass of our metal sheet, which is 3.0 kg multiplied by the acceleration due to gravity which is 9.8 m per second squared plus 10 newtons per meter means that's the units on the value for 10. And don't forget that it's multiplied by some value X all multiplied by sine of 15 degrees. So the reason why this equation is super important is because we need to use this equation to plug in our different values of X to help us plot our graph. So we need to note that we are gonna be calculating and to make things simple, we will be looking at these two graphs here. OK. So paying attention just on their left side right now. So we will calculate X equals 00.51 0.0 and 1.5 and 2.0 m. So when we plug in the value X equals 0 m, so when we use our kinetic friction equation. And then when we plug in the value for zero, for where X is, we will get that FK is equal to 10.29 newtons. And using the same methodology for 0.5 m, we will get 10.74 newtons. And for 1.0 m, we will get 11.19 newtons. And for 1.5 m, we'll get 11.65 newtons. And for 2.0 m, we will get 12.10 newtons. So when we plot these points on our graph, we will get the following orange line. So this is our kinetic friction is on our vertical axis and our x of m the distance is our horizon, sorry our horizontal axe. And as we could see, we have a ST a steady linear graph trending upwards. So now this is what we look like when we plot it in figure two. So when we plot this information in figure two, without the line of applied force F, it will give us the graph on the right, which is, as you could see is our skeleton graph from figure three has our line. So when we take this and we plot this into our figure three, we will see that just slightly above 10, it will be trending upwards. We have this nice nice diagonal line that's trending upwards and that's it. That's how we solve for this problem. So it'd be helpful for a problem like this. If you might use like graphing software or like your graphing calculator to help you solve this or if you had some graph paper, so you could plot this out on your own. But it's really easy, you just have to make sure you set up the equation properly. And once the equation is set up properly, then you'll be able to solve for this super easy. But this, here we'll put a highlighted box is our final answer. This is our correct and final answer and that's it. So just as long as you get the equation set up correctly, plugging in your values of X will be super easy. And you want to do a like at least you know, 3 to 5 numbers just to make sure you got a nice little line plot or a nice little graph that you could plot and that's it. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

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