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Unformatted text preview: Patel (ppp285) 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 . 6 kg, the coefficient of kinetic friction between the block and the wall be = 0 . 63, and = 69 . Suppose F = 72 N. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 6 kg F 6 9 k =0 . 63 Find the force of friction. Correct answer: 16 . 2555 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 = 16 . 2555 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 cos = mg + F sin 2. F cos = mg 3. F sin = mg F sin 4. F cos = mg F cos 5. F cos = mg F sin 6. F sin = mg + F sin 7. F cos = mg + F cos 8. F sin = mg F cos 9. F sin = mg + F cos correct 10. F sin = mg 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 3.40 kg block is pushed along the ceiling with a constant applied force of 89.0 N that acts at an angle of 61.0 with the horizontal. The block accelerates to the right at 6.30 m/s 2 . The acceleration of gravity is 9 . 81 m / s 2 . 3 . 4 kg 8 9 N 6 1 6 . 3 m / s 2 Patel (ppp285) HW04 TSOI (58160) 2 What is the coefficient of kinetic friction between the block and the ceiling? Correct answer: 0 . 488412. 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 = 3 . 40 kg = 61 . F applied = 89 N a x = 6 . 30 m g = 9 . 81 m / s 2 Solution: F applied,x = (89 N) cos61 = 43 . 1481 N F applied,y = (89 N) sin61 = 77 . 8412 N The normal force is F n = F applied,y mg = 77 . 8412 N (3 . 4 kg)(9 . 81 m / s 2 ) = 44 . 4872 N From the horizontal motion, ma x = F applied,x k F n k = F applied,x ma x F n = 43 . 1481 N (3 . 4 kg)(6 . 3 m / s 2 ) 44 . 4872 N = 0 . 488412 004 (part 1 of 3) 10.0 points The suspended 2 . 9 kg mass on the right is moving up, the 1 . 5 kg mass slides down the ramp, and the suspended 8 . 4 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 14 ....
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- Spring '08