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You and three friends stand at the corners of a square whose sides are 8.0 $\mathrm{m}$ long in the middle of the gym floor, as shown in Fig. $\mathrm{E} 7.31 .$ You take your physics book and push it from one person to the other. The book has a mass of $1.5 \mathrm{kg},$ and the coefficient of kinetic

friction between the book and the floor is $\mu_{\mathrm{k}}=0.25 .$ (a) The book slides from you to Beth and then from Beth to Carlos, along the lines connecting these people. What is the work done by friction during this displacement? (b) You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement? (c) You slide the book to Kim, who then slides it back to you. What is the total work done by friction during this motion of the book? (d) Is the friction force on the book conservative or nonconservative? Explain.

(a) $-29.4 \mathrm{J}$

(b)$-42 \mathrm{J}$

(c) $-29.4 \mathrm{J}$

(d) From part (c), since the total work done during a complete round trip is not zero, therefore the friction force on the book is nonconservative.

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Taylor J.

January 9, 2021

How isn’t is a correct? I have this same problem and the number above isn’t given.

Taylor J.

January 9, 2021

Woah sorry my phone messed that up but how is A correct?*^

you and three friends stand at the corners of a square whose size are eight meters long. In this problem, you'll be taking your physics book and pushing around the square. The book has a massive 1.5 kilograms, and the coefficient of kinetic friction between the book and the floor is 0.25 In the first part of this problem, the book slides from you to Beth and then from Beth to Carlos, and the problem wants you to find the work done by friction during this movement of the book. So I'm here. Recall that the formula for kinetic friction is f k equals Myu f n, where F n is the normal force. However, since the book is only moving horizontally, normal force will be equal to F G, which is equal o m. G. So if we put that back into the kinetic force equation, we get F K. It's equal to new M G. But here we're not trying to find force. We're trying to find the work done by the friction, so work is equal to negative force of friction times. The district's at the book travels here. The negative sign is because the direction of the friction force is opposite of the direction of movement. So if we look everything into the work equation, we have negative new times. M times G times d We're do used again distance of travel. So for part A, we have negative Europe one Choose five from you 1.5 kilograms for the mass of a book 9.8 years per second. Squared for junior in 16 meters for a deed since I moved from you to Beth and then bet to Carlos. This will give us about negative 58 point a Jules. And that would be the work done by Friction Force as the tramples from you to Beth and then from back to Carlos. So now I'm part B. We have the book sliding from you to Carlos along the diagonal of the square. Let's go ahead and I'll be this diagram and put it next page so that we have more space. All right, now you want to look at the books movement along the diagonal of the of the square from you to Carlos again, they work. The creation will be negative mu and G times the distance that it travels. So here we have 0.25 for the new 1.5 kilograms for the mass, 9.8 meters per second squared Fergie and here d will be the diagonal which from trigonometry, remember is eight times rad, too meters. So if we crunch all the numbers in a calculator, we'll end up with negative 41 point six Jules for the work done in part B by the Friction force in part C, you slide the book to Kim and then Kim slides it back to you. So again, I'm going to copy the diagram so that we have more space when you get rid of the diagonal. And we're only looking for the friction for us on the book when it slides from you to Kim and then came back to you. Now notice that the distance that it travels is 16 16 meters, eight meters took him and eight meters back to you. This is exactly the same distance that the book travels in part A. So rather than going through everything, I'll just write the equation first negative from you and G do you Since mu does not change from part A. M does not change from part A. Neither does G, and neither does Dean because the distance is the same. We get the same answer as part A. We're just 58.8. Jules now Part D asked if the force the friction force on the book, is conservative. So for conservative forces, when an object moves in a path such that it ends at the same point as it starts. If the four stone on the object is zero, then it's conservative, so W equals zero means conservative. If the work is not zero, then it's not a conservative here. In this case, since the book goes to Cannes and then comes back to you, the starting and ending point is the same. However, we noticed that the work done is not zero. Yes, the friction force on the book as it slides from you to Kim and then came back to you is non conservative. And if you do the calculation for the work on any other half that starts and ends at the same point, you'll find that the work is always non zero. That's the The friction force on the book is always non conservative,

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