This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hernandez (ejh742) – oldhomework 15 – Turner – (58120) 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 θ with respect to the horizontal. The coefficient of kinetic friction is μ . The box stops after sliding a distance x . m μ k v θ How far does the box slide? 1. x = v 2 g ( μ sin θ + cos θ ) 2. x = v 2 2 g μ cos θ 3. x = v 2 2 g ( μ cos θ − sin θ ) correct 4. x = v 2 2 g ( μ sin θ + cos θ ) 5. x = v 2 g ( μ sin θ − 2 cos θ ) 6. x = v 2 2 g (sin θ − μ cos θ ) 7. x = v 2 2 g ( μ sin θ − cos θ ) 8. x = v 2 g (sin θ − μ cos θ ) 9. x = v 2 2 g ( μ cos θ + sin θ ) 10. x = v 2 2 g sin θ Explanation: The net force on the block parallel to the incline is F net = F mg sin θ − F f , where F f is the friction force. Thus, Newton’s equation for the block reads ma = mg sin θ − F f = mg sin θ − μ N = mg (sin θ − μ cos θ ) a = g (sin θ − μ cos θ ) , where N = mg cos θ . 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 θ − μ cos θ ) = v 2 2 g ( μ cos θ − sin θ ) . 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 = (2 m / s 3 ) t 3 − (5 m / s 2 ) t 2 . What is the velocity of the particle at any time t ? 1. v = (6 m / s 2 ) t 2 − (4 m / s) t 2. v = (12 m / s 2 ) t 2 − (6 m / s) t 3. v = (15 m / s 2 ) t 2 − (4 m / s) t 4. v = (6 m / s 2 ) t 2 − (10 m / s) t correct 5. v = (9 m / s 2 ) t 2 − (6 m / s) t 6. v = (24 m / s 2 ) t 2 − (8 m / s) t 7. v = (27 m / s 2 ) t 2 − (4 m / s) t 8. v = (18 m / s 2 ) t 2 − (14 m / s) t 9. v = (27 m / s 2 ) t 2 − (8 m / s) t hernandez (ejh742) – oldhomework 15 – Turner – (58120) 2 10. v = (21 m / s 2 ) t 2 − (10 m / s) t Explanation: The velocity of the particle is v = dx dt = d dt bracketleftbig (2 m / s 3 ) t 3 − (5 m / s 2 ) t 2 bracketrightbig = (6 m / s 2 ) t 2 − (10 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 − (10 m) correct 2. a = (12 m / s) t − (4 m) 3. a = (36 m / s) t − (12 m) 4. a = (30 m / s) t − (4 m) 5. a = (42 m / s) t − (6 m) 6. a = (18 m / s) t − (6 m) 7. a = (48 m / s) t − (12 m) 8. a = (24 m / s) t − (14 m) 9. a = (36 m / s) t − (14 m) 10. a = (24 m / s) t − (18 m) Explanation: The acceleration of the particle is a = dv dt = d dt bracketleftbig (6 m / s 2 ) t 2 − (10 m / s) t bracketrightbig = (12 m / s) t − (10 m)...
View
Full
Document
This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Mass, Work

Click to edit the document details