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Unformatted text preview: markowitz (am45362) – HW05 – distler – (56295) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. There are various ways to do part 2 of the ”Wind and Pendulum” problem. The easiest is to find the angle at which the three forces on the ball (tension, gravity and F) add up to zero. In part 1 of ”Pulling Two Masses”, you are asked about the ”work done against fric tion”. This is very funny phraseology. The friction does negative work, and what you are being askedfor is the magnitude (a positive number) of the work done by friction. 001 10.0 points A weight lifter lifts a mass m at constant speed to a height h in time t . How much work W is done by the weight lifter? 1. W = mg h correct 2. W = mg h t 3. W = mg 4. W = mh 5. W = mg ht Explanation: By conservation of energy, the work done to the mass is the change in its energy. Since Δ v = 0, only the gravitational potential en ergy changes, so the work is W = Δ U = mg h . 002 10.0 points A spring with spring constant of 35 N / m is stretched 0 . 18 m from its equilibrium position. How much work must be done to stretch it an additional 0 . 1 m? Correct answer: 0 . 805 J. Explanation: The new displacement is 0 . 28 m, so W = Δ U = 1 2 k bracketleftBig x 2 x 2 bracketrightBig = 1 2 (35 N / m) bracketleftBig (0 . 28 m) 2 (0 . 18 m) 2 bracketrightBig = 0 . 805 J . 003 (part 1 of 2) 10.0 points A 0 . 7 kg bead slides on a curved wire, starting from rest at point A as shown in the figure. The segment from A to B is frictionless, and the segment from B to C is rough. The point A is at height 8 . 9 m and the point C is at height 0 . 3 m with respect to point B. The acceleration of gravity is 9 . 8 m / s 2 . A 8.9 m B .3 m C Find the speed of the bead at B. Correct answer: 13 . 2076 m / s. Explanation: Given : m = 0 . 7 kg and h = 8 . 9 m . Choose the zero level for potential energy at the level of B. Between A and B, K A + U A = K B + U B 0 + mg H = 1 w mv 2 + 0 v = radicalbig 2 g H = radicalBig 2(9 . 8 m / s 2 )(8 . 9 m) = 13 . 2076 m / s . 004 (part 2 of 2) 10.0 points If the bead comes to rest at C, find the change in mechanical energy due to friction as it moves from B to C. markowitz (am45362) – HW05 – distler – (56295) 2 Correct answer: 58 . 996 J. Explanation: Given : h = 0 . 3 m Choose the starting point at B, the zero po tential energy level at B, as before, and the end point at C. Then the energy loss is equal to the work done by the nonconservative forces and is W nc = K f K i + U f U i = 0 1 2 mv 2 B + mg h = (0 . 7 kg) (13 . 2076 m / s) 2 2 + (0 . 7 kg) (9 . 8 m / s 2 ) (0 . 3 m) = 58 . 996 J ....
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This note was uploaded on 11/29/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas.
 Spring '08
 Kaplunovsky
 Physics

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