This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: quiz 02 – GAUTHIER, JOSEPH – Due: Oct 17 2007, 10:00 pm 1 Question 1, chap 6, sect 1. part 1 of 1 10 points A box weighing 740 N is pushed along a horizontal floor at constant velocity with a force of 280 N parallel to the floor. What is the coefficient of kinetic friction between the box and the floor? 1. . 378378 correct 2. . 390625 3. . 402778 4. . 415385 5. . 428571 6. . 442623 7. . 457627 8. . 472727 9. . 490909 10. . 508475 Explanation: Let : W = 740 N and F app = 280 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 = 740 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 = 280 N 740 N = . 378378 . 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 25 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient μ ). The radius of the track is 51 m. The acceleration of gravity is 9 . 8 m / s 2 . 51 m 25 kg θ If the kinetic energy of the block at the bottom of the track is 6000 J, what is the work done against friction? 1. 5882 . 4 J 2. 6094 . 6 J 3. 6296 J 4. 6495 J correct 5. 6704 . 8 J 6. 6950 J 7. 7184 . 8 J 8. 7470 J 9. 7813 . 2 J 10. 8084 . 8 J Explanation: W = W f + K W f = mg R K = (25 kg) (9 . 8 m / s 2 ) (51 m) (6000 J) = 6495 J . Question 3, chap 8, sect 5. part 1 of 1 10 points A rain cloud contains 6.73 × 10 7 kg of water vapor. The acceleration of gravity is 9 . 81 m / s 2 . How long would it take for a 2.53 kW pump to raise the same amount of water to the cloud’s altitude of 2.61 km? quiz 02 – GAUTHIER, JOSEPH – Due: Oct 17 2007, 10:00 pm 2 1. 5 . 83819 × 10 8 s 2. 6 . 02114 × 10 8 s 3. 6 . 21185 × 10 8 s 4. 6 . 40478 × 10 8 s 5. 6 . 60288 × 10 8 s 6. 6 . 81089 × 10 8 s correct 7. 7 . 02603 × 10 8 s 8. 7 . 24475 × 10 8 s 9. 7 . 50985 × 10 8 s 10. 7 . 7741 × 10 8 s Explanation: Basic Concepts: W = P Δ t W = Fd cos θ = Fd = mgd since θ = 0 ◦ ⇒ cos θ = 1. Given: m = 6 . 73 × 10 7 kg P = 2 . 53 kW d = 2 . 61 km g = 9 . 81 m / s 2 Solution: Δ t = W P = mgd P = (6 . 73 × 10 7 kg)(9 . 81 m / s 2 )(2610 m) 2530 W = 6 . 81089 × 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 . 7 kg 47 kg a Find the acceleration of the mass on the left....
View
Full
Document
This homework help was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Gilbert

Click to edit the document details