SOL19 - Thevalingam Donald Homework 19 Due 6:00 pm Inst...

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Thevalingam, Donald – Homework 19 – Due: Feb 19 2007, 6:00 pm – Inst: Eslami 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Lazy Kids! You need to do the problems faster. 001 (part 1 of 3) 10 points A block is pushed against the spring with spring constant k (located on the left-hand side of the track) and compresses the spring a distance 4 . 7 cm from its equilibrium position (as shown in the figure below). The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track except for a small rough area on a horizontal section of the track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9 . 81 m / s 2 . μ =0 . 5 1 m 574 g 2 m 2 m 9 . 81 m / s 2 v k 4 . 7 cm What is the spring constant k ? Correct answer: 5098 . 18 N / m. Explanation: Let : g = 9 . 81 m / s 2 , m = 0 . 574 kg , x = 2 m , v x = v , = 1 m , d = 0 . 047 m , and μ = 0 . 5 . μ =0 . 5 m h x g 3 . 13 m / s 5 . 1 kN / m d Basic Concepts: Conservation of Me- chanical Energy U i = U f + K f + W (1) since v i = 0 m/s. K = 1 2 m v 2 (2) U s = 1 2 k d 2 (3) W = μ m g ‘ . (4) Choosing the point where the block leaves the track as the origin of the coordinate system, Δ x = v x Δ t Δ y = - 1 2 g Δ t 2 since a x i = 0 m/s 2 and v y i = 0 m/s. Solution: Using Eqs. 1, 2, 3, and 4, we have 1 2 m v 2 x = 1 2 k d 2 - μ m g ‘ (5) k = m d 2 v 2 x + 2 μ g ‘ (6) h = - 1 2 g t 2 (7) x = v x t . (8) Using Eq. 7 and substituting t = x v x from Eq. 8, we have h = - 1 2 g x v x 2 , so v 2 x = - g x 2 2 h . (9)

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Thevalingam, Donald – Homework 19 – Due: Feb 19 2007, 6:00 pm – Inst: Eslami 2 Using Eq. 6 and substituting v 2 x from Eq. 8, we have k = m d 2 v 2 x + 2 μ g ‘ (10) = m d 2 - g x 2 2 h + 2 μ g ‘ = 0 . 574 kg (0 . 047 m) 2 × - (9 . 81 m / s 2 ) (2 m) 2 2 ( - 2 m) + 2 (0 . 5) (9 . 81 m / s 2 ) (1 m) = 5098 . 18 N / m . 002 (part 2 of 3) 10 points What is the speed v of the block when it leaves the track? Correct answer: 3 . 13209 m / s. Explanation: Using Eq. 5, we have v 2 x = k d 2 m - 2 μ g ‘ v x = r k d 2 m - 2 μ g ‘ = (5098 . 18 N / m) (0 . 047 m) 2 (0 . 574 kg) - 2 (0 . 5) (9 . 81 m / s 2 ) (1 m) 1 / 2 = 3 . 13209 m / s . Alternate Solution: At Δ y = h = - 2 m (below the jump off height), h = - 1 2 g t 2 (11) t = s - 2 h g .
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