{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics Test2 Solutions

# Physics Test2 Solutions - quiz 02 VENNES ROSS Due 10:00 pm...

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

quiz 02 – VENNES, ROSS – Due: Oct 17 2007, 10:00 pm 1 Question 1, chap 6, sect 1. part 1 of 1 10 points A box weighing 660 N is pushed along a horizontal floor at constant velocity with a force of 240 N parallel to the floor. What is the coefficient of kinetic friction between the box and the floor? 1. 0 . 277778 2. 0 . 287671 3. 0 . 298507 4. 0 . 308824 5. 0 . 318841 6. 0 . 328767 7. 0 . 338983 8. 0 . 352113 9. 0 . 363636 correct 10. 0 . 375 Explanation: Let : W = 660 N and F app = 240 N . vector F app vector f k vector N m vector W Because the box is moving with constant velocity, its acceleration is zero and the net foce acting on it is zero. Applying summationdisplay F y = 0 to the box, N - W = 0 N = W = 660 N . Applying summationdisplay F x = 0 to the box, F app - f k = 0 f k = F app μ k N = F app μ k = F app N = 240 N 660 N = 0 . 363636 . Question 2, chap 7, sect 1. part 1 of 1 10 points Starting from rest at a height equal to the radius of the circular track, a block of mass 11 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient μ ). The radius of the track is 36 m. The acceleration of gravity is 9 . 8 m / s 2 . 36 m 11 kg θ If the kinetic energy of the block at the bottom of the track is 1200 J, what is the work done against friction? 1. 2294 . 6 J 2. 2367 . 2 J 3. 2444 J 4. 2520 . 4 J 5. 2599 . 2 J 6. 2680 . 8 J correct 7. 2764 . 8 J 8. 2852 . 4 J 9. 2951 J 10. 3044 . 8 J Explanation: W = W f + K W f = m g R - K = (11 kg) (9 . 8 m / s 2 ) (36 m) - (1200 J) = 2680 . 8 J . Question 3, chap 8, sect 5. part 1 of 1 10 points A rain cloud contains 5.67 × 10 7 kg of water vapor. The acceleration of gravity is 9 . 81 m / s 2 . How long would it take for a 2.09 kW pump to raise the same amount of water to the cloud’s altitude of 2.64 km?

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

View Full Document
quiz 02 – VENNES, ROSS – Due: Oct 17 2007, 10:00 pm 2 1. 5 . 44503 × 10 8 s 2. 5 . 65828 × 10 8 s 3. 5 . 83819 × 10 8 s 4. 6 . 02114 × 10 8 s 5. 6 . 21185 × 10 8 s 6. 6 . 40478 × 10 8 s 7. 6 . 60288 × 10 8 s 8. 6 . 81089 × 10 8 s 9. 7 . 02603 × 10 8 s correct 10. 7 . 24475 × 10 8 s Explanation: Basic Concepts: W = P Δ t W = Fd cos θ = Fd = mgd since θ = 0 cos θ = 1. Given: m = 5 . 67 × 10 7 kg P = 2 . 09 kW d = 2 . 64 km g = 9 . 81 m / s 2 Solution: Δ t = W P = mgd P = (5 . 67 × 10 7 kg)(9 . 81 m / s 2 )(2640 m) 2090 W = 7 . 02603 × 10 8 s Question 4, chap 5, sect 6. part 1 of 1 10 points Consider the following system of two masses and two pulleysl The pulleys that are massless and frictionless, and the mass on the left accelerates upward while the mass on the right accelerates downward. The acceleration of gravity is 9 . 8 m / s 2 . 5 kg 25 kg a Find the acceleration of the mass on the left. 1. 6 . 10492 m / s 2 2. 6 . 32258 m / s 2 3. 6 . 53333 m / s 2 correct 4. 6 . 7375 m / s 2 5. 7 m / s 2 6. 7 . 22105 m / s 2 7. 7 . 49412 m / s 2 8. 7 . 84 m / s 2 9. 8 . 12683 m / s 2 10. 8 . 4 m / s 2 Explanation: Let : m 1 = 5 kg m 2 = 25 kg and g = 9 . 8 m / s 2 . m 1 m 2 1 2 T T T a Let a denote the upward acceleration of the m 1 mass on the left and let T be the tension of the long string attached to the m 1 . To understand the motion of the right mass m 2 , consider the free-body diagram for the right pulley: The pulley is massless, so regardless of its acceleration all forces acting on the pulley must balance; the tension of the short string connecting the right pulley to the m 2 mass must be 2 T rather than T . At the same time, the unstretchability of the long string requires the pulley — and hence the m 2 mass
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