hw7 - karna (pk4534) HW 7 coker (58245) 1 This print-out...

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Unformatted text preview: karna (pk4534) HW 7 coker (58245) 1 This print-out should have 23 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 single conservative force F ( x ) = bx + a acts on a 2 . 71 kg particle, where x is in meters, b = 4 . 76 N / m and a = 5 . 06 N. As the particle moves along the x axis from x 1 = 0 . 442 m to x 2 = 4 . 05 m, calculate the work done by this force. Correct answer: 56 . 8295 J. Explanation: The work done by the conservative force is W = integraldisplay F x dx = integraldisplay ( bx + a ) dx = 1 2 bx 2 + ax vextendsingle vextendsingle vextendsingle x 2 x 1 = 1 2 (4 . 76 N / m) x 2 + (5 . 06 N) x vextendsingle vextendsingle vextendsingle 4 . 05 m . 442 m = 59 . 5309 J 2 . 70149 J = 56 . 8295 J . 002 (part 2 of 3) 10.0 points Calculate the change in the potential energy of the particle. Correct answer: 56 . 8295 J. Explanation: From conservation of energy K + U = 0 , we obtain U = K = W = 56 . 8295 J . 003 (part 3 of 3) 10.0 points Calculate the particles initial kinetic energy at x 1 if its final speed at x 2 is 17 . 7 m / s. Correct answer: 367 . 678 J. Explanation: The change in the kinetic energy is K = K f K i K i = K f K K i = 1 2 mv 2 f K = 1 2 (2 . 71 kg)(17 . 7 m / s) 2 56 . 8295 J = 367 . 678 J . 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 mg k 2. x = v m k g 3. x = v 2 2 m 4. x = v radicalbigg k m 5. x = v mk g 6. x = v radicalbigg m k correct 7. x = v radicalBigg k mg 8. x = v 2 2 g 9. x = v k g m 10. x = v radicalbigg mg k karna (pk4534) HW 7 coker (58245) 2 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 . Anyone who checks to see if the units are correct should get this problem correct. 005 (part 1 of 2) 10.0 points A bead slides without friction around a loop- the-loop. The bead is released from position A at a height y A (which may vary) from the bottom of the loop-the-loop which has a ra- dius r . The acceleration of gravity is g . y A r C B A What is the instantaneous kinetic energy K C at C so that the bead would press the track with an upward force F press = 1 2 mg ? 1. K C = mg r 2. K C = 1 2 mg r 3. K C = 2 3 mg r 4. K C = 3 4 mg r correct 5. K C = 1 4 mg r 6. K C = 2 mg r 7. K C = 3 2 mg r Explanation: Applying Newtons third law at C , the mag- nitude of the upward force by which the bead presses the track is the same as the magnitude of the downward normal force on the bead by the track, i.e. , as given in the question...
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This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw7 - karna (pk4534) HW 7 coker (58245) 1 This print-out...

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