# 15 (5) - hernandez(ejh742 oldhomework 15 Turner(58120 This...

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hernandez (ejh742) – oldhomework 15 – Turner – (58120) 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 0 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 0 θ How far does the box slide? 1. x = v 2 0 g ( μ sin θ + cos θ ) 2. x = v 2 0 2 g μ cos θ 3. x = v 2 0 2 g ( μ cos θ sin θ ) correct 4. x = v 2 0 2 g ( μ sin θ + cos θ ) 5. x = v 2 0 g ( μ sin θ 2 cos θ ) 6. x = v 2 0 2 g (sin θ μ cos θ ) 7. x = v 2 0 2 g ( μ sin θ cos θ ) 8. x = v 2 0 g (sin θ μ cos θ ) 9. x = v 2 0 2 g ( μ cos θ + sin θ ) 10. x = v 2 0 2 g sin θ Explanation: The net force on the block parallel to the incline is F net = F m g sin θ F f , where F f is the friction force. Thus, Newton’s equation for the block reads m a = m g sin θ F f = m g sin θ μ N = m g (sin θ μ cos θ ) a = g (sin θ μ cos θ ) , where N = m g cos θ . To find the distance the block slides down the incline, use v 2 = v 2 0 + 2 a ( x x 0 ) , valid for a body moving with a constant accel- eration. Since x 0 = 0 and v f = 0 (the block stops), we get x = v 2 0 2 a = v 2 0 2 g (sin θ μ cos θ ) = v 2 0 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

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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 = d x 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 = d v dt = d dt bracketleftbig (6 m / s 2 ) t 2 (10 m / s) t bracketrightbig = (12 m / s) t (10 m) . 004 (part 3 of 4) 10.0 points What is the power delivered to the particle at any time t ? 1. P = 2 m t bracketleftBig (225 m 2 / s 6 ) t 2 (315 m 2 / s 5 ) t + (98 m 2 / s 4 ) bracketrightBig 2. P = 2 m t bracketleftBig (225 m 2 / s 6 ) t 2 (180 m 2 / s 5 ) t + (32 m 2 / s 4 ) bracketrightBig 3. P = 4 m t bracketleftBig (162 m 2 / s 6 ) t 2 (189 m 2 / s 5 ) t + (49 m 2 / s 4 ) bracketrightBig 4. P = 18 m t bracketleftBig (49 m 2 / s 6 ) t
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