phucHW7

# phucHW7 - Quach Phuc – Homework 7 – Due Oct 6 2006 5:00...

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Unformatted text preview: Quach, Phuc – Homework 7 – Due: Oct 6 2006, 5:00 pm – Inst: Mark Rupright 1 This print-out should have 19 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 The potential energy between two atoms in a particular molecule has the form U ( x ) = 2 . 4 x 8- 5 . 6 x 4 where the units of x are length and the num- bers 2 . 4 and 5 . 6 have appropriate units so that U ( x ) has units of energy. What is the equilibrium separation of the atoms (that is the distance at which the force between the atoms is zero)? Correct answer: 0 . 962195 . Explanation: The force as a function of position can be calculated as the negative of the derivative of U ( x ); F ( x ) =- dU dx = (19 . 2) x 9- (22 . 4) x 5 Thus, the condition F = 0 is satisfied when x = 4 r 19 . 2 22 . 4 = 0 . 962195 . keywords: 002 (part 1 of 1) 10 points Water flows over a section of Niagara Falls at a rate of 1 . 44 × 10 6 kg / s and falls 70 m. The acceleration of gravity is 9 . 8 m / s 2 . How many 80 W bulbs can be lit with this power? Correct answer: 1 . 2348 × 10 7 . Explanation: Let : r = 1 . 44 × 10 6 kg / s , and h = 70 m . The power of the water is equal to the change in potential energy per unit time, so P = U t = mg h t = m t g h = (1 . 44 × 10 6 kg / s)(9 . 8 m / s 2 )(70 m) = 9 . 8784 × 10 8 W . The power required to light n = 80 W light bulbs is P = n (80 W) . Solving for n , n = P 80 W = 9 . 8784 × 10 8 W 80 W = 1 . 2348 × 10 7 . keywords: 003 (part 1 of 1) 10 points Ittakes 5 . 56 J of work to stretcha Hooke’s-law spring 13 . 9 cm from its unstressed length. How much the extra work is required to stretch it an additional 14 . 5 cm? Correct answer: 17 . 6504 J. Explanation: We begin by determining the value of k from the energy equation solved for k , k = 2 U x 2 = 2(5 . 56 J) (0 . 139 m) 2 = 575 . 54 J / m 2 We can now determine the energy at the ad- ditional displacement, U add = 1 2 k x 2 add = 1 2 575 . 54 J / m 2 (0 . 139 m + 0 . 145 m) 2 = 23 . 2104 J The extra work required is just the difference in energy between the two displacements. W = Δ U = U add- U = 23 . 2104 J- 5 . 56 J = 17 . 6504 J keywords: Quach, Phuc – Homework 7 – Due: Oct 6 2006, 5:00 pm – Inst: Mark Rupright 2 004 (part 1 of 1) 10 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 r mg k 2. x = v 2 2 m 3. x = v mg k 4. x = v m k g 5. x = v s k mg 6. x = v mk g 7. x = v r m k correct 8. x = v 2 2 g 9. x = v k g m 10. x = v r k m 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 r m k .....
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## This note was uploaded on 11/30/2008 for the course PHY 2048 taught by Professor Guzman during the Fall '08 term at FAU.

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phucHW7 - Quach Phuc – Homework 7 – Due Oct 6 2006 5:00...

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