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Unformatted text preview: fullwood (tf5349) – Homework Conservation of Energy – Catala – (49033) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 49 kg pole vaulter running at 10 m / s vaults over the bar. Her speed when she is above the bar is 0 . 7 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole. Correct answer: 5 . 07704 m. Explanation: Given : m = 49 kg , v i = 10 m / s , and v f = 0 . 7 m / s . Applying conservation of mechanical energy, 1 2 mv 2 i = mg h f + 1 2 mv 2 f v 2 i = 2 g h f + v 2 f h f = v 2 i − v 2 f 2 g = (10 m / s) 2 − (0 . 7 m / s) 2 2 (9 . 8 m / s 2 ) = 5 . 07704 m . 002 10.0 points The ball launcher in a pinball machine has a spring that has a force constant of 1 . 47 N / cm. The surface on which the ball moves is in clined 12 . 4 ◦ with respect to the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 12 . 4 ◦ 1 . 47 N / cm 1 . 47 N / cm 1 . 47 N / cm 1 . 47 N / cm 1 . 47 N / cm 1 . 47 N / cm 1 . 47 N / cm 1 . 47 N / cm v = 0 v 4 . 21 cm 12 . 4 ◦ If the spring is initially compressed 4 . 21 cm, find the launching speed of a 0 . 179 kg ball when the plunger is released. Friction and the mass of the plunger are negligible. Correct answer: 1 . 13065 m / s. Explanation: k v = 0 v Δ x θ PE = 0 h Given : k = 1 . 47 N / cm , m = 0 . 179 kg , θ = 12 . 4 ◦ , and Δ x = 4 . 21 cm . Consider U g = 0 at the point of compres sion. For the compression Δ x of the spring, the ball rises h = Δ x sin θ while the spring returns to its equilibrium position. The initial energy is ( K + U g + U s ) i = 1 2 k (Δ x ) 2 and the final energy is ( K + U g + U s ) f = 1 2 mv 2 f + mg h. Equating the energies, 1 2 mv 2 f + mg Δ x sin θ = 1 2 k (Δ x ) 2 mv 2 f = k (Δ x ) 2 − 2 mg Δ x sin θ v 2 f = k (Δ x ) 2 m − 2 g Δ x sin θ fullwood (tf5349) – Homework Conservation of Energy – Catala – (49033) 2 = (1 . 47 N / cm) (4 . 21 cm) 2 . 179 kg parenleftbigg 1 m 100 cm parenrightbigg − (2) ( 9 . 8 m / s 2 ) (4 . 21 cm) (sin12 . 4 ◦ ) · parenleftbigg 1 m 100 cm parenrightbigg = 1 . 27836 m 2 / s 2 . Thus v f = radicalBig 1 . 27836 m 2 / s 2 = 1 . 13065 m / s . 003 (part 1 of 2) 10.0 points A 0 . 48 kg bead slides on a curved wire, start ing from rest at point A as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . A 6.1 m B 1.66 m C If the wire is frictionless, find the speed of the bead at B. Correct answer: 10 . 9343 m / s. Explanation: Given : m = 0 . 48 kg and h A = 6 . 1 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 A = 1 2 mv 2 + 0 v = radicalbig 2 g h A = radicalBig 2(9 . 8 m / s 2 )(6 . 1 m) = 10 . 9343 m / s 004 (part 2 of 2) 10.0 points Find the speed of the bead at C....
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 Spring '10
 PRO.KIROSHIMA
 Conservation Of Energy, Energy, Potential Energy, Work, Correct Answer, Conservation of Mechanical Energy

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