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Unformatted text preview: 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 = 0 . 43, and = 62 . Suppose F = 79 N. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 5 kg F 6 2 k =0 . 43 Find the force of friction. Correct answer: 15 . 9479 N. Explanation: Recall that f = N . From summationdisplay F = 0 in the horizontal direction, one can see that N = F cos . Hence the force of friction is f = N = F cos = 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 = mg 2. F cos = mg F cos 3. F cos = mg + F sin 4. F cos = mg 5. F sin = mg F cos 6. F cos = mg F sin 7. F sin = mg + F cos correct 8. F sin = mg F sin 9. F cos = mg + F cos 10. F sin = mg + F sin Explanation: m F v mg f For constant velocity, acceleration is zero. Hence summationdisplay F y = 0 = F sin mg F cos . 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 9 N 6 7 m / s 2 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 F applied,y = F applied sin F y,net = F applied,y mg F n = 0 F x,net = ma x = F applied,x F k F k = k F n m F a Given: m = 4 . 80 kg = 60 . F applied = 90 N a x = 7 . 00 m g = 9 . 81 m / s 2 Solution: F applied,x = (90 N) cos60 = 45 N F applied,y = (90 N) sin60 = 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 k F n 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 ....
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This note was uploaded on 11/03/2010 for the course PHYSICS 58160 taught by Professor Tsoi during the Spring '10 term at University of Texas at Austin.
- Spring '10