{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# quiz1010 - Version 035/AACAD quiz10-10 Demkov(59910 This...

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

Version 035/AACAD – quiz10-10 – Demkov – (59910) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Pretend you are on a planet similar to Earth where the acceleration of gravity is approxi- mately 10 m / s 2 . The coefficient of kinetic friction between the plane and the block is 0 . 4. A block of mass 25 kg lies on an inclined plane, as shown below. The horizontal and vertical supports for the plane have lengths of 20 m and 15 m, respectively. 25 kg μ = 0 . 4 F 20 m 15 m The magnitude of the force vector F necessary to pull the block up the plane with constant speed is most nearly 1. bardbl vector F bardbl ≈ 200 N 2. bardbl vector F bardbl ≈ 236 N 3. bardbl vector F bardbl ≈ 258 N 4. bardbl vector F bardbl ≈ 126 N 5. bardbl vector F bardbl ≈ 150 N 6. bardbl vector F bardbl ≈ 172 N 7. bardbl vector F bardbl ≈ 244 N 8. bardbl vector F bardbl ≈ 488 N 9. bardbl vector F bardbl ≈ 230 N correct 10. bardbl vector F bardbl ≈ 270 N Explanation: Let : x = 20 m , y = 15 m , m = 25 kg , g 10 m / s 2 , and a = 0 m / s 2 . s θ x y s = radicalbig x 2 + y 2 = radicalBig (20 m) 2 + (15 m) 2 = 25 m , sin θ = y s = 15 25 , and cos θ = x s = 20 25 . Consider the free body diagram for the block m g sin θ N = m g cos θ μ N F m g Parallel to the ramp, F net = F g bardbl + F k F = m a = 0 , F g bardbl = m g sin θ , F k = μ k N , so F = F g bardbl + F k . (1) Perpendicular to the ramp, F net = N F g = 0 , so N = m g cos θ . (2)

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

View Full Document
Version 035/AACAD – quiz10-10 – Demkov – (59910) 2 Using Eqs. 1 and 2, parallel to the ramp, we have F = F g bardbl + F k = m g (sin θ + μ k cos θ ) = (25 kg) (10 m / s 2 ) bracketleftbigg 15 25 + (0 . 4) 20 25 bracketrightbigg = 230 N . keywords: 002 10.0 points Two identical massless springs are hung from a horizontal support. A block of mass 4 . 4 kg is suspended from the pair of springs, as shown. The acceleration of gravity is 9 . 8 m / s 2 . k k 4 . 4 kg When the block is in equilibrium, each spring is stretched an additional 0 . 37 m. The force constant of each spring is most nearly 1. k 37 N / m 2. k 110 N / m 3. k 51 N / m 4. k 56 N / m 5. k 58 N / m correct 6. k 66 N / m 7. k 41 N / m 8. k 22 N / m 9. k 140 N / m 10. k 82 N / m Explanation: Let : m = 4 . 4 kg , x = 0 . 37 m
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern