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r3s9R54gXg67_1210016105_jwk572

# r3s9R54gXg67_1210016105_jwk572 - homework 05 KIM JI Due Oct...

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homework 05 – KIM, JI – Due: Oct 3 2007, 4:00 am 1 Question 1, chap 8, sect 1. part 1 of 1 10 points A 8 . 2 kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 2 . 8 kg mass. The acceleration of gravity is 9 . 8 m / s 2 . 6 . 1 m ω 8 . 2 kg 2 . 8 kg Use conservation of energy to determine the final speed of the first mass after it has fallen (starting from rest) 6 . 1 m. Correct answer: 7 . 66114 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 8 . 2 kg , m 2 = 2 . 8 kg , and = 6 . 1 m . Consider the free body diagrams 8 . 2 kg 2 . 8 kg T T m 1 g m 2 g a a Let the figure represent the initial config- uration of the pulley system (before m 1 falls down). From the conservation of energy K i + U i = K f + U f 0 + m 1 g ℓ = m 2 g ℓ + 1 2 m 1 v 2 + 1 2 m 2 v 2 ( m 1 m 2 ) g ℓ = 1 2 ( m 1 + m 2 ) v 2 Therefore v = radicalBigg ( m 1 m 2 ) ( m 1 + m 2 ) 2 g ℓ = bracketleftbigg 8 . 2 kg 2 . 8 kg 8 . 2 kg + 2 . 8 kg × 2 (9 . 8 m / s 2 )(6 . 1 m) bracketrightbigg 1 / 2 = 7 . 66114 m / s . Question 2, chap 8, sect 1. part 1 of 2 10 points A 0 . 4 kg bead slides on a curved wire, start- ing 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 . 3 m and the point C is at height 1 . 8 m with respect to point B. The acceleration of gravity is 9 . 8 m / s 2 . A 8.3 m B 1.8 m C Find the speed of the bead at B. Correct answer: 12 . 7546 m / s (tolerance ± 1 %). Explanation: Given : m = 0 . 4 kg and h = 8 . 3 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

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homework 05 – KIM, JI – Due: Oct 3 2007, 4:00 am 2 0 + m g H = 1 w m v 2 + 0 v = radicalbig 2 g H = radicalBig 2(9 . 8 m / s 2 )(8 . 3 m) = 12 . 7546 m / s . Question 3, chap 8, sect 1. part 2 of 2 10 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: 25 . 48 J (tolerance ± 1 %). Explanation: Given : h = 1 . 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 . 4 kg) (12 . 7546 m / s) 2 2 + (0 . 4 kg) (9 . 8 m / s 2 ) (1 . 8 m) = 25 . 48 J . Question 4, chap 8, sect 1. part 1 of 1 10 points A block starts at rest and slides down 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 . 3 1 . 4 m 420 g h 2 . 3 m 3 . 81 m 9 . 81 m / s 2 v At what height h above the ground is the block released? Correct answer: 4 . 29784 m (tolerance ± 1 %). Explanation: Let : x = 3 . 81 m , g = 9 . 81 m / s 2 , m = 420 g , μ = 0 . 3 , = 1 . 4 m , h 2 = 2 . 3 m , h = h 1 h 2 , and v x = v .
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r3s9R54gXg67_1210016105_jwk572 - homework 05 KIM JI Due Oct...

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