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Unformatted text preview: hernandez (ejh742) oldhomework 14 Turner (58120) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A 6 . 1 kg particle moves along the x axis under the influence of a single conservative force. If the work done on the particle is 50 J as it moves from x 1 = 2 . 9 m to x 2 = 6 . 9 m, find the change in its kinetic energy. Correct answer: 50 J. Explanation: The change in the kinetic energy is equal to the work done: K = W = 50 J . 002 (part 2 of 3) 10.0 points Find the change in its potential energy. Correct answer:- 50 J. Explanation: Since K + U = 0 , then U =- K =- 50 J . 003 (part 3 of 3) 10.0 points Find its speed at x 2 = 6 . 9 m if it starts from rest. Correct answer: 4 . 04888 m / s. Explanation: Using K = m ( v 2 end- v 2 init ) 2 we obtain v = radicalbigg 2 K m = radicalBigg 2(50 J) 6 . 1 kg = 4 . 04888 m / s . 004 10.0 points A block of mass m slides on a horizontal frictionless table with an initial speed v . 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 = 0 How much is the spring compressed x from its natural length? 1. x = v m k g 2. x = v 2 2 g 3. x = v radicalbigg m k correct 4. x = v 2 2 m 5. x = v mg k 6. x = v mk g 7. x = v radicalbigg k m 8. x = v k g m 9. x = v radicalBigg k mg 10. x = v radicalbigg mg k Explanation: Total energy is conserved (no friction). The spring is compressed by a distance x from its natural length, so 1 2 mv 2 = E i = E f = 1 2 k x 2 , or x 2 = m k v 2 , therefore x = v radicalbigg m k . hernandez (ejh742) oldhomework 14 Turner (58120) 2 Anyone who checks to see if the units are correct should get this problem correct. 005 10.0 points A bead slides without friction around a loop- the-loop. The bead is released from a height of 27 m from the bottom of the loop-the-loop which has a radius 9 m....
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