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Unformatted text preview: karna (pk4534) – HW 7 – coker – (58245) 1 This printout should have 23 questions. Multiplechoice 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 Δ K + Δ U = 0 , we obtain Δ U = − Δ K = − W = − 56 . 8295 J . 003 (part 3 of 3) 10.0 points Calculate the particle’s 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 Δ K = K f − K i K i = K f − Δ K K i = 1 2 mv 2 f − Δ 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 μ = 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 theloop. The bead is released from position A at a height y A (which may vary) from the bottom of the looptheloop 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 Newton’s 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.
 Spring '08
 Turner
 Physics, Force

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