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: lenoir (wml297) oldhomework 15 Turner (58220) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A box of mass m with an initial velocity of v slides down a plane, inclined at with respect to the horizontal. The coefficient of kinetic friction is . The box stops after sliding a distance x . m k v How far does the box slide? 1. x = v 2 2 g sin 2. x = v 2 2 g cos 3. x = v 2 2 g ( cos + sin ) 4. x = v 2 2 g (sin cos ) 5. x = v 2 2 g ( cos sin ) correct 6. x = v 2 g (sin cos ) 7. x = v 2 g ( sin + cos ) 8. x = v 2 2 g ( sin + cos ) 9. x = v 2 2 g ( sin cos ) 10. x = v 2 g ( sin 2 cos ) Explanation: The net force on the block parallel to the incline is F net = F mg sin F f , where F f is the friction force. Thus, Newtons equation for the block reads ma = mg sin F f = mg sin N = mg (sin cos ) a = g (sin cos ) , where N = mg cos . To find the distance the block slides down the incline, use v 2 = v 2 + 2 a ( x x ) , valid for a body moving with a constant accel eration. Since x = 0 and v f = 0 (the block stops), we get x = v 2 2 a = v 2 2 g (sin cos ) = v 2 2 g ( cos sin ) . 002 (part 1 of 4) 10.0 points A particle of mass m moves along the x axis. Its position varies with time according to x = (6 m / s 3 ) t 3 (8 m / s 2 ) t 2 . What is the velocity of the particle at any time t ? 1. v = (6 m / s 2 ) t 2 (12 m / s) t 2. v = (6 m / s 2 ) t 2 (18 m / s) t 3. v = (12 m / s 2 ) t 2 (14 m / s) t 4. v = (9 m / s 2 ) t 2 (8 m / s) t 5. v = (12 m / s 2 ) t 2 (16 m / s) t 6. v = (24 m / s 2 ) t 2 (12 m / s) t 7. v = (18 m / s 2 ) t 2 (16 m / s) t correct 8. v = (21 m / s 2 ) t 2 (18 m / s) t 9. v = (6 m / s 2 ) t 2 (10 m / s) t lenoir (wml297) oldhomework 15 Turner (58220) 2 10. v = (6 m / s 2 ) t 2 (4 m / s) t Explanation: The velocity of the particle is v = dx dt = d dt bracketleftbig (6 m / s 3 ) t 3 (8 m / s 2 ) t 2 bracketrightbig = (18 m / s 2 ) t 2 (16 m / s) t . 003 (part 2 of 4) 10.0 points What is the acceleration of the particle at any time t ? 1. a = (12 m / s) t (16 m) 2. a = (24 m / s) t (18 m) 3. a = (36 m / s) t (6 m) 4. a = (42 m / s) t (8 m) 5. a = (42 m / s) t (16 m) 6. a = (54 m / s) t (14 m) 7. a = (36 m / s) t (16 m) correct 8. a = (48 m / s) t (12 m) 9. a = (12 m / s) t (18 m) 10. a = (54 m / s) t (10 m) Explanation: The acceleration of the particle is a = dv dt = d dt bracketleftbig (18 m / s 2 ) t 2 (16 m / s) t bracketrightbig = (36 m / s) t (16 m) ....
View
Full
Document
This note was uploaded on 04/15/2010 for the course PHY 12343 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

Click to edit the document details