Homework 6 Solutions - homework 06 ALIBHAI ZAHID Due 4:00...

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homework 06 – ALIBHAI, ZAHID – Due: Feb 28 2007, 4:00 am 1 Question 1 part 1 of 1 10 points The figure below shows a rough semicircu- lar track whose ends are at a vertical height h . A block placed at point P at one end of the track is released from rest and slides past the bottom of the track. h P What height on the far side of the track does the block reach if the coefficient of kinetic friction is 0 . 3 . 1. h = 0 . 3 h 2 2. h = h 3. h = 0 . 3 h 2 π 4. h = 0 . 7 h 2 π 5. h = 0 . 7 h 6. h = 0 . 3 h 7. It is between zero and h ; the exact height depends on how much energy is lost to fric- tion. correct 8. h = 0 . 7 h 2 Explanation: Because the surface of the track is rough, the block will lose some energy to friction ( W < 0). Use the work energy theorem E f E i = W, we know that the height h to which the block rises on the other side of the track should be E f E i = W m g h m g h = W < 0 m g h < m g h h < h Therefore, 0 < h < h . Question 2 part 1 of 1 10 points A 9 . 8 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 . 9 kg mass. The acceleration of gravity is 9 . 8 m / s 2 . 6 . 2 m ω 9 . 8 kg 2 . 9 kg Use conservation of energy to determine the final speed of the first mass after it has fallen (starting from rest) 6 . 2 m. Correct answer: 8 . 12543 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 9 . 8 kg , m 2 = 2 . 9 kg , and = 6 . 2 m . Consider the free body diagrams 9 . 8 kg 2 . 9 kg T T m 1 g m 2 g a a
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homework 06 – ALIBHAI, ZAHID – Due: Feb 28 2007, 4:00 am 2 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 9 . 8 kg 2 . 9 kg 9 . 8 kg + 2 . 9 kg × 2 (9 . 8 m / s 2 )(6 . 2 m) bracketrightbigg 1 / 2 = 8 . 12543 m / s . Question 3 part 1 of 2 10 points Consider the loop-the-loop setup, where the mass is released from rest at C with height h . The mass is sliding along a frictionless track, and the radius of the loop is R . h R A B C D m Determine the height h such that the cen- tripetal acceleration at A is a c = 2 g . 1. h = 5 R 2. h = 4 1 2 R 3. h = 4 R 4. h = 3 1 2 R 5. h = 2 1 2 R 6. h = 3 R correct 7. h = 2 R Explanation: The centripetal force at point A is given by F c = m a c = m (2 g ) = m v 2 A R = K A = 1 2 m v 2 A = m g R . From the conservation of energy, U C + K C = U A + K A , or m g h + 0 = m g (2 R ) + 1 2 m v 2 A = 3 m g R = h = 3 R . Question 4 part 2 of 2 10 points Consider a new situation where h = 4 R . The speed at the bottom of the track is 1. bardbl vectorv bardbl = radicalbig 5 g R . 2. bardbl vectorv bardbl = radicalbig 3 g R . 3. bardbl vectorv bardbl = radicalbig 6 g R . 4. bardbl vectorv bardbl = radicalbig 9 g R .
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