hw 14 - brown (twb493) hw 14 turner (57340) This print-out...

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brown (twb493) – hw 14 – turner – (57340) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 3) 10.0 points A 6 . 3 kg particle moves along the x axis under the in±uence oF a single conservative Force. IF the work done on the particle is 42 J as it moves From x 1 = 1 . 8 m to x 2 = 3 . 8 m, fnd the change in its kinetic energy. Correct answer: 42 J. Explanation: The change in the kinetic energy is equal to the work done: Δ K = W = 42 J . 002 (part 2 of 3) 10.0 points ²ind the change in its potential energy. Correct answer: - 42 J. Explanation: Since Δ K + Δ U = 0 , then Δ U = - Δ K = - 42 J . 003 (part 3 of 3) 10.0 points ²ind its speed at x 2 = 3 . 8 m iF it starts From rest. Correct answer: 3 . 65148 m / s. Explanation: Using K = m ( v 2 end - v 2 init ) 2 we obtain v = r K m = R 2(42 J) 6 . 3 kg = 3 . 65148 m / s . 004 10.0 points A block oF mass m slides on a horizontal Frictionless table with an initial speed v 0 . It then compresses a spring oF Force constant k and is brought to rest. v m k m μ = 0 How much is the spring compressed x From its natural length? 1. x = v 0 k g m 2. x = v 0 mg k 3. x = v 0 m k g 4. x = v 2 0 2 m 5. x = v 0 mk g 6. x = v 0 r k m 7. x = v 2 0 2 g 8. x = v 0 r m k correct 9. x = v 0 R k 10. x = v 0 r k Explanation: Total energy is conserved (no Friction). The spring is compressed by a distance x From its natural length, so 1 2 mv 2 0 = E i = E f = 1 2 k x 2 , or x 2 = m k v 2 0 , thereFore x = v 0 r m k .
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This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw 14 - brown (twb493) hw 14 turner (57340) This print-out...

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