oldhw17 - hinojosa(jlh3938 – oldhomework 17 – Turner...

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Unformatted text preview: hinojosa (jlh3938) – oldhomework 17 – Turner – (58185) 1 This print-out should have 14 questions. Multiple-choice 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 2 g ( μ sin θ + cos θ ) 2. x = v 2 2 g ( μ cos θ − sin θ ) correct 3. x = v 2 g ( μ sin θ − 2 cos θ ) 4. x = v 2 2 g (sin θ − μ cos θ ) 5. x = v 2 2 g sin θ 6. x = v 2 2 g μ cos θ 7. x = v 2 g (sin θ − μ cos θ ) 8. x = v 2 g ( μ sin θ + cos θ ) 9. x = v 2 2 g ( μ sin θ − cos θ ) 10. x = v 2 2 g ( μ cos θ + 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 = (7 m / s 3 ) t 3 − (6 m / s 2 ) t 2 . What is the velocity of the particle at any time t ? 1. v = (6 m / s 2 ) t 2 − (10 m / s) t 2. v = (24 m / s 2 ) t 2 − (12 m / s) t 3. v = (6 m / s 2 ) t 2 − (14 m / s) t 4. v = (27 m / s 2 ) t 2 − (14 m / s) t 5. v = (18 m / s 2 ) t 2 − (14 m / s) t 6. v = (18 m / s 2 ) t 2 − (18 m / s) t 7. v = (27 m / s 2 ) t 2 − (18 m / s) t 8. v = (12 m / s 2 ) t 2 − (6 m / s) t 9. v = (21 m / s 2 ) t 2 − (12 m / s) t correct hinojosa (jlh3938) – oldhomework 17 – Turner – (58185) 2 10. v = (9 m / s 2 ) t 2 − (12 m / s) t Explanation: The velocity of the particle is v = dx dt = d dt bracketleftbig (7 m / s 3 ) t 3 − (6 m / s 2 ) t 2 bracketrightbig = (21 m / s 2 ) t 2 − (12 m / s) t . 003 (part 2 of 4) 10.0 points What is the acceleration of the particle at any time t ? 1. a = (42 m / s) t − (16 m) 2. a = (54 m / s) t − (12 m) 3. a = (24 m / s) t − (14 m) 4. a = (30 m / s) t − (18 m) 5. a = (12 m / s) t − (6 m) 6. a = (42 m / s) t − (12 m) correct 7. a = (12 m / s) t − (10 m) 8. a = (36 m / s) t − (12 m) 9. a = (48 m / s) t − (12 m) 10. a = (48 m / s) t − (6 m) Explanation: The acceleration of the particle is a = dv dt = d dt bracketleftbig (21 m / s 2 ) t 2 − (12 m / s) t bracketrightbig = (42 m / s) t − (12 m) ....
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This note was uploaded on 04/08/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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oldhw17 - hinojosa(jlh3938 – oldhomework 17 – Turner...

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