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# hw 5 - sousse(res2468 HW 5 Kleinman(58225 This print-out...

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sousse (res2468) – HW 5 – Kleinman – (58225) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This HW covers Chs. 7 and 8. Note that multiple choice questions with fewer than 8 answer choices have been reduced in value to only 5 points. 001 10.0 points A woman weight lifter can lift a 130 lb weight from the floor to a stand 3 . 2 ft off the ground. What is the total work done by the woman? Correct answer: 564 . 096 J. Explanation: Let : F = 130 lb and d = 3 . 2 ft . W = F d = (130 lb) (3 . 2 ft) × 1 . 356 J ft · lb = 564 . 096 J . 002 10.0 points A catcher “gives” with a baseball when catch- ing it. If the baseball exerts a force of 464 N on the glove such that the glove is displaced 13.2 cm, how much work is done by the ball? Correct answer: 61 . 248 J. Explanation: Basic Concept: W net = F net d cos θ = F net d since θ = 0 cos θ = 1. Given: F net = 464 N d = 13 . 2 cm Solution: W net = F net d = (464 N)(0 . 132 m) = 61 . 248 J 003 (part 1 of 2) 10.0 points A 2 kg object is given a displacement Δ vectors = (3 m)ˆ ı + (1 m)ˆ + ( 4 m) ˆ k along a straight line. During the displacement, a constant force vector F = (4 N)ˆ ı + ( 5 N)ˆ + (5 N) ˆ k acts on the object. Find the work done by vector F for this displace- ment. Correct answer: 13 J. Explanation: Let : m = 2 kg , Δ vectors = (3 m)ˆ ı + (1 m)ˆ + ( 4 m) ˆ k , and vector F = (4 N)ˆ ı + ( 5 N)ˆ + (5 N) ˆ k . The work is W = vector F · Δ vectors = F x s x + F y s y + F y s z = (4 N) (3 m) + ( 5 N) (1 m) + (5 N) ( 4 m) = 13 J . 004 (part 2 of 2) 10.0 points Find the component of vector F in the direction of this displacement. Correct answer: 2 . 54951 N. Explanation: W = vector F · Δ vectors = vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle | Δ vectors | cos θ . The component of vector F in the direction of Δ vectors is vextenddouble vextenddouble vextenddouble vector F vextenddouble vextenddouble vextenddouble cos θ = W bardbl Δ vectors bardbl , so F cos θ = W radicalBig s 2 x + s 2 y + s 2 z = 1 J radicalbig (3 m) 2 + (1 m) 2 + ( 4 m) 2 = 2 . 54951 N .

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sousse (res2468) – HW 5 – Kleinman – (58225) 2 005 10.0 points A block of mass m is pushed a horizontal dis- tance D from position A to position B, along a horizontal plane with friction coefficient μ . Then m is pushed from B to A. If the force pushing m from A to B is vector P , and the force pushing m from B to A is vector P , what is the total work done by friction? 1. 2 ( P μ m g ) D 2. 2 ( μ m g P ) D 3. +2 μ m g D 4. 0 5. 2 μ m g D correct Explanation: From A to B, the work done by friction is μ m g D . From B to A, the work done by frition is μ m g D too since the work done by kinetic friction is always negative. So the total work done by friction is equal to the sum of these two works, and we get 2 μ m g D .
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