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# HS7 - Adams Russell Homework 7 Due 11:00 pm Inst Opyrchal H...

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Adams, Russell – Homework 7 – Due: Oct 27 2006, 11:00 pm – Inst: Opyrchal, H 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. 001 (part 1 of 1) 10 points A block of mass m slides on a horizontal frictionless table with an initial speed v 0 . It then compresses a spring of force constant k and is brought to rest. The acceleration of gravity is 9 . 8 m / s 2 . v m k m μ = 0 How much is the spring compressed x from its natural length? 1. x = v 0 m g k 2. x = v 0 m k g 3. x = v 2 0 2 m 4. x = v 0 k g m 5. x = v 0 r m g k 6. x = v 0 r m k correct 7. x = v 2 0 2 g 8. x = v 0 s k m g 9. x = v 0 r k m 10. x = v 0 m k g Explanation: Total energy is conserved (no friction). The spring is compressed by a distance x from its natural length, so 1 2 m v 2 0 = E i = E f = 1 2 k x 2 , or x 2 = m k v 2 0 , therefore x = v 0 r m k . Anyone who checks to see if the units are correct should get this problem correct. keywords: 002 (part 1 of 2) 10 points A bead slides without friction around a loop- the-loop. The bead is released from a height 11 . 7 m from the bottom of the loop-the-loop which has a radius 4 m. The acceleration of gravity is 9 . 8 m / s 2 . 11 . 7 m 4 m A What is its speed at point A ? Correct answer: 8 . 51587 m / s. Explanation: Let : m = 4 g , R = 4 m , and h = 11 . 7 m . From conservation of energy, we have K i + U i = K f + U f 0 + m g h = m v 2 2 + m g (2 R ) v 2 = 2 g ( h - 2 R ) . Therefore v = p 2 g ( h - 2 R ) = r 2 (9 . 8 m / s 2 ) h 11 . 7 m - 2 (4 m) i = 8 . 51587 m / s .

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Adams, Russell – Homework 7 – Due: Oct 27 2006, 11:00 pm – Inst: Opyrchal, H 2 003 (part 2 of 2) 10 points How large is the normal force on it at point A if its mass is 4 g? Correct answer: 0 . 03332 N. Explanation: The centripetal force is equal to the normal force due to the track and gravity, F c = N + m g , so N = m v 2 R - m g = (4 g)(0 . 001 kg / g) × (8 . 51587 m / s) 2 4 m - 9 . 8 m / s 2 = 0 . 03332 N . keywords: 004 (part 1 of 1) 10 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 . 3 1 . 4 m 462 g h 2 m 4 . 08 m 9 . 81 m / s 2 v At what height h above the ground is the block released?
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HS7 - Adams Russell Homework 7 Due 11:00 pm Inst Opyrchal H...

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