HW 7-solutions

HW 7-solutions - morris (gmm643) HW 7 Coker (56625) This...

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morris (gmm643) – HW 7 – Coker – (56625) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. This assignment covers the general defni- tion oF potential energy For a conservative Force, general types oF conservation oF energy examples (with 2 or more Forces doing work), potential energy diagrams, and the concept oF binding energy. 001 10.0 points A synthetic rubber band resists being stretched a distance x From equilibrium with a Force v F b ( x ) = ˆ ıb x 2 , where b is a constant. What is the potential energy U b ( x ) associ- ated with this elastic band? 1. b x 3 3 2. 2 b x 3. + b x 2 2 4. + b x 3 3 correct 5. b x 2 2 6. +2 b x 7. zero Explanation: By defnition Δ U = W . The work we would have to do against the Force v F b to stretch the band is i b x 2 dx = b x 3 3 . 002 10.0 points In a certain region oF space, a particle expe- riences a potential energy U ( x ) = A x 2 + B , where A and B are constants. What Force F gives rise to this potential energy? 1. ˆ i p A x 2 P 2. ˆ i p A x 2 P 3. ˆ i p A x + B x P 4. There is no Force, since the slope oF the line is the constant B A . 5. ˆ i p A x + B x P 6. ˆ i p A x P 7. ˆ i p 2 A x 3 P correct 8. ˆ i p A x P 9. ˆ i p 2 A x 3 P Explanation: The Force is F = dU dx ˆ i = ˆ i p 2 A x 3 P . 003 10.0 points A block oF mass m 1 is attached to a horizon- tal spring oF Force constant k and to a spring oF negligible mass. The string runs over a mass- less, Frictionless pulley to a hanging block oF mass m 2 . Initially, the entire system is at rest and the spring is unstretched.

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morris (gmm643) – HW 7 – Coker – (56625) 2 m 1 m 2 k m 1 m 2 If mass m 1 slides on a horizontal frictionless surface, what is the speed v of the mass m 2 when it has fallen a distance downward from its rest position? 1. Zero, since the spring will stop it from falling. 2. r 2 m 2 g ℓ k ℓ 2 m 1 + m 2 correct 3. R ± ± ² m 2 g ℓ 1 2 k ℓ 2 m 1 + m 2 4. r 2 m 2 g ℓ k ℓ 2 m 1 m 2 5. R ± ± ² m 2 g ℓ 1 2 k ℓ 2 2 ( m 1 + m 2 ) 6. r 2 m 2 g ℓ + k ℓ 2 m 1 m 2 7. r 2 m 2 g ℓ + k ℓ 2 m 1 + m 2 Explanation: Let the gravitational potential energy have a value of 0 at the end state when the system has moved a distance . Applying conserva- tion of energy, E 0 = E f U 0 + K 0 = U f + K 1 ,f + K 2 ,f m 2 g ℓ + 0 = 1 2 k ℓ 2 + 1 2 m 1 v 2 + 1 2 m 2 v 2 2 m 2 g ℓ = k ℓ 2 + ( m 1 + m 2 ) v 2 2 m 2 g ℓ k ℓ 2 = ( m 1 + m 2 ) v 2 v 2 = 2 m 2 g ℓ k ℓ 2 m 1 + m 2 v = r 2 m 2 g ℓ k ℓ 2 m 1 + m 2 . 004 10.0 points You put a ball in a spring gun and compress the spring a distance d . The ball is Fred straight up and reaches height above the gun.
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This note was uploaded on 11/17/2010 for the course PHY 56630 taught by Professor Coker during the Spring '10 term at University of Texas.

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HW 7-solutions - morris (gmm643) HW 7 Coker (56625) This...

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