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Unformatted text preview: markowitz (am45362) – HW05 – distler – (56295) 1 This print-out should have 21 questions. Multiple-choice 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 asked-for 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 non-conservative 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