# HW 4 - yang(ey942 HW04 TSOI(58160 1 This print-out should...

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: yang (ey942) 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 = 3 . 1 kg, the coefficient of kinetic friction between the block and the wall be = 0 . 58, and = 60 . Suppose F = 67 N. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 1 kg F 6 k =0 . 58 Find the force of friction. Correct answer: 19 . 43 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 = 19 . 43 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 + F cos 3. F sin = mg + F cos correct 4. F cos = mg 5. F sin = mg + F sin 6. F cos = mg F sin 7. F sin = mg 8. F sin = mg F sin 9. F sin = mg F cos 10. F cos = mg F cos 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 3.80 kg block is pushed along the ceiling with a constant applied force of 87.0 N that acts at an angle of 65.0 with the horizontal. The block accelerates to the right at 6.70 m/s 2 . The acceleration of gravity is 9 . 81 m / s 2 . 3 . 8 kg 8 7 N 6 5 6 . 7 m / s 2 yang (ey942) HW04 TSOI (58160) 2 What is the coefficient of kinetic friction between the block and the ceiling? Correct answer: 0 . 272013. 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 . 80 kg = 65 . F applied = 87 N a x = 6 . 70 m g = 9 . 81 m / s 2 Solution: F applied,x = (87 N) cos65 = 36 . 7678 N F applied,y = (87 N) sin65 = 78 . 8488 N The normal force is F n = F applied,y mg = 78 . 8488 N (3 . 8 kg)(9 . 81 m / s 2 ) = 41 . 5708 N From the horizontal motion, ma x = F applied,x k F n k = F applied,x ma x F n = 36 . 7678 N (3 . 8 kg)(6 . 7 m / s 2 ) 41 . 5708 N = 0 . 272013 004 (part 1 of 3) 10.0 points The suspended 3 kg mass on the right is mov- ing up, the 2 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 . 16 ....
View Full Document

## This note was uploaded on 11/21/2010 for the course PHY 58160 taught by Professor Tsio during the Fall '10 term at University of Texas.

### Page1 / 11

HW 4 - yang(ey942 HW04 TSOI(58160 1 This print-out should...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online