{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW4 Solution - salazar(ggs332 HW04 TSOI(58160 This...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
salazar (ggs332) – HW04 – TSOI – (58160) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points As shown in the figure, a block is pushed up against the wall. Let the mass of the block be m = 2 . 5 kg, the coefficient of kinetic friction between the block and the wall be μ = 0 . 43, and θ = 62 . Suppose F = 79 N. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 5 kg F 62 μ k = 0 . 43 Find the force of friction. Correct answer: 15 . 9479 N. Explanation: Recall that f = μ N . From summationdisplay F = 0 in the horizontal direction, one can see that N = F cos θ . Hence the force of friction is f = μ N = μ F cos θ = 15 . 9479 N . 002 (part 2 of 2) 10.0 points The force, F , which keeps the block mov- ing upwards with a constant velocity satisfies which equations? 1. F sin θ = μ m g 2. F cos θ = m g μ F cos θ 3. F cos θ = m g + μ F sin θ 4. F cos θ = μ m g 5. F sin θ = m g μ F cos θ 6. F cos θ = m g μ F sin θ 7. F sin θ = m g + μ F cos θ correct 8. F sin θ = m g μ F sin θ 9. F cos θ = m g + μ F cos θ 10. F sin θ = m g + μ F sin θ Explanation: m F θ v m g f μ For constant velocity, acceleration is zero. Hence summationdisplay F y = 0 = F sin θ m g μ F cos θ . The first term is the applied upward force, the second term is the weight of the block, and the third term is the frictional force. 003 10.0 points A 4.80 kg block is pushed along the ceiling with a constant applied force of 90.0 N that acts at an angle of 60.0 with the horizontal. The block accelerates to the right at 7.00 m/s 2 . The acceleration of gravity is 9 . 81 m / s 2 . 4 . 8 kg 90 N 60 μ 7 m / s 2
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
salazar (ggs332) – HW04 – TSOI – (58160) 2 What is the coefficient of kinetic friction between the block and the ceiling? Correct answer: 0 . 369479. Explanation: Basic Concepts: F applied,x = F applied cos θ F applied,y = F applied sin θ F y,net = F applied,y mg F n = 0 F x,net = ma x = F applied,x F k F k = μ k F n m F θ μ a Given: m = 4 . 80 kg θ = 60 . 0 F applied = 90 N a x = 7 . 00 m g = 9 . 81 m / s 2 Solution: F applied,x = (90 N) cos 60 = 45 N F applied,y = (90 N) sin 60 = 77 . 9423 N The normal force is F n = F applied,y mg = 77 . 9423 N (4 . 8 kg)(9 . 81 m / s 2 ) = 30 . 8543 N From the horizontal motion, ma x = F applied,x μ k F n μ k = F applied,x ma x F n = 45 N (4 . 8 kg)(7 m / s 2 ) 30 . 8543 N = 0 . 369479 004 (part 1 of 3) 10.0 points The suspended 2 . 7 kg mass on the right is moving up, the 2 . 5 kg mass slides down the ramp, and the suspended 7 . 1 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 17 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 5 kg μ = 0 . 17 23 7 . 1 kg 2 . 7 kg What is the acceleration of the three block system? Correct answer: 3 . 97228 m / s 2 . Explanation: Let : m 1 = 2 . 7 kg , m 2 = 2 . 5 kg , m 3 = 7 . 1 kg , and θ = 23 .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern