extra_credit 02-solutions

extra_credit 02-solutions - marout (ssm979) – extra...

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Unformatted text preview: marout (ssm979) – extra credit 02 – turner – (56725) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points On planet Tehar, the free-fall acceleration is the same as as that of Earth, but there is also a strong downward electric field that is uniform close to the planet’s surface. A 5 . 92 kg ball having a charge of 5 . 92 μ C is thrown upward at a speed of 10 . 3 m / s, and it hits the ground after an interval of 4 . 92 s. The acceleration of gravity is 9 . 8 m / s 2 . What is the potential difference between the starting point and the top point of the trajectory? Correct answer:- 71111 . 2 kV. Explanation: Let : q = 5 . 92 μ C = 5 . 92 × 10- 6 C , v i = 10 . 3 m / s , g = 9 . 8 m / s 2 , m = 5 . 92 kg , and t = 4 . 92 s . For the full flight, Δ y = v i t + 1 2 a t 2 = 0 a =- 2 v i t . For the upward part of the flight, y f = 0, so v i 2 + 2 a h = 0 v i 2- 4 v i h t = 0 4 v i h = v i 2 t h = v i t 4 . From Newton’s second law, a = ∑ F y m =- mg- q E m =- 2 v i t g + g E m = 2 v i t E = m q parenleftbigg 2 v i t- g parenrightbigg . Thus V = E h = parenleftbigg m q parenrightbigg bracketleftbigg 2 v i t- g bracketrightbigg parenleftbigg v i t 4 parenrightbigg = 5 . 92 kg 5 . 92 × 10- 6 C × bracketleftbigg 2 (10 . 3 m / s) 4 . 92 s- 9 . 8 m / s 2 bracketrightbigg × (10 . 3 m / s) (4 . 92 s) 4 · 1 kV 1000 V =- 71111 . 2 V . 002 10.0 points Two identical metal blocks resting on a frictionless horizontal surface are connected by a light metal spring having constant of 173 N / m and unstretched length of 0 . 2 m. A total charge of Q is slowly placed on the system causing the spring to stretch to an equilibrium length of 0 . 5 m. Determine the value of Q , assuming that all the charge resides on the blocks and the blocks can be treated as point charges. The permittivity of free space is 8 . 85419 × 10- 12 C 2 / N / m 2 . Correct answer: 7 . 59911 × 10- 5 C. Explanation: Let : k = 173 N / m and Δ x = 0 . 5 m- . 2 m = 0 . 3 m . The system is in static equilibrium, so that the sum of the forces on each block is zero. And since the metal blocks are identical, they get equal charges of Q/ 2. For a spring stretch of Δ x , the sum of the forces on either block is written as summationdisplay F =- k Δ x + parenleftbigg Q 2 parenrightbigg 2 4 π ǫ x 2 , marout (ssm979) – extra credit 02 – turner – (56725) 2 since by symmetry the charge Q will be di- vided evenly between the blocks by symmetry. Solving for Q in the above expression, Q = radicalbig k Δ x 16 π ǫ x 2 = radicalBig (173 N / m) (0 . 3 m) 16 π ǫ (0 . 5 m) 2 = 7 . 59911 × 10- 5 C ....
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This note was uploaded on 10/24/2010 for the course PHY 56735 taught by Professor Turner,j during the Fall '10 term at University of Texas.

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extra_credit 02-solutions - marout (ssm979) – extra...

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