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# hw7 - al-ameri(aoa434 Hwk07 Stokes(19102 This print-out...

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al-ameri (aoa434) – Hwk07 – Stokes – (19102) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is ONLINE HOMEWORK No 7. It is due before 7.00 am on MONDAY 16 March, Abu Dhabi Time (9.00 pm on Sunday in Texas) 001 (part 1 of 3) 10.0 points A single conservative force F ( x ) = b x + a acts on a 4 . 5 kg particle, where x is in meters, b = 3 . 23 N / m and a = 3 . 06 N. As the particle moves along the x axis from x 1 = 1 . 06 m to x 2 = 4 . 58 m, calculate the work done by this force. Correct answer: 42 . 8335 J. Explanation: The work done by the conservative force is W = integraldisplay F x dx = integraldisplay ( b x + a ) dx = 1 2 b x 2 + a x vextendsingle vextendsingle vextendsingle x 2 x 1 = 1 2 (3 . 23 N / m) x 2 + (3 . 06 N) x vextendsingle vextendsingle vextendsingle 4 . 58 m 1 . 06 m = 47 . 8917 J - 5 . 05821 J = 42 . 8335 J . 002 (part 2 of 3) 10.0 points Calculate the change in the potential energy of the particle. Correct answer: - 42 . 8335 J. Explanation: From conservation of energy Δ K + Δ U = 0 , we obtain Δ U = - Δ K = - W = - 42 . 8335 J . 003 (part 3 of 3) 10.0 points Calculate the particle’s initial kinetic energy at x 1 if its final speed at x 2 is 16 . 3 m / s. Correct answer: 554 . 969 J. Explanation: The change in the kinetic energy is Δ K = K f - K i K i = K f - Δ K K i = 1 2 m v 2 f - Δ K = 1 2 (4 . 5 kg)(16 . 3 m / s) 2 - 42 . 8335 J = 554 . 969 J . 004 (part 1 of 2) 10.0 points A 0 . 2 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 . 8 m with respect to point B. The acceleration of gravity is 9 . 8 m / s 2 . A 8.9 m B .8 m C Find the speed of the bead at B. Correct answer: 13 . 2076 m / s. Explanation: Given : m = 0 . 2 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 + m g H = 1 w m v 2 + 0 v = radicalbig 2 g H = radicalBig 2(9 . 8 m / s 2 )(8 . 9 m) = 13 . 2076 m / s .

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al-ameri (aoa434) – Hwk07 – Stokes – (19102) 2 005 (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. Correct answer: - 15 . 876 J. Explanation: Given : h = 0 . 8 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 m v 2 B + m g h - 0 = (0 . 2 kg) (13 . 2076 m / s) 2 2 + (0 . 2 kg) (9 . 8 m / s 2 ) (0 . 8 m) = - 15 . 876 J . 006 (part 1 of 3) 10.0 points A block starts at rest and slides down a fric- tionless 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 . 2 1 . 1 m 497 g h 2 . 1 m 4 . 92 m 9 .
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