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Kaplunovsky - exam 03

# Kaplunovsky - exam 03 - midterm 03 – – Due Apr 4 2007...

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Unformatted text preview: midterm 03 – – Due: Apr 4 2007, 11:00 pm 1 Question 1, chap 8, sect 1. part 1 of 1 10 points Consider a bungee cord of unstretched length L = 35 m. When the cord is stretched to L > L it behaves like a spring and its ten- sion follows the Hooke’s law T = k ( L − L ). But unlike a spring, the cord folds instead of becoming compressed when the distance be- tween its ends is less than the unstretched length: For L < L the cord has zero tension and zero elastic energy. To test the cord’s reliability, one end is tied to a high bridge (height H = 103 m above the surface of a river) and the other end is tied to a steel ball of weight mg = 100 kg × 9 . 8 m / s 2 . The ball is dropped off the bridge with zero initial speed. Fortunately, the cord works and the ball stops in the air 18 m above the water — and then the cord pulls it back up. Calculate the cord’s ‘spring’ constant k . For simplicity, neglect the cord’s own weight and inertia as well as the air drag on the ball and the cord. 1. 62 . 5767 N / m 2. 64 . 5511 N / m 3. 66 . 64 N / m correct 4. 68 . 8976 N / m 5. 71 . 2727 N / m 6. 73 . 9253 N / m 7. 76 . 3298 N / m 8. 78 . 707 N / m 9. 81 . 2334 N / m 10. 83 . 9001 N / m Explanation: In the absence of air drag and other resis- tive forces, there are only two forces acting on the ball – the gravity force mvectorg and the cord’s tension T , which are both conserva- tive. Therefore, the system (the ball plus the cord) has conserved mechanical energy E mech = K + U grav + U cord = const , where K = mv 2 2 , U grav = mg h, U cord = braceleftbigg 1 2 k ( L − L ) 2 for L > L , for L < L . When the ball is dropped off the bridge, it has zero initial speed (hence K = 0) while the cord is folded rather than stretched and hence U cord = 0 as well. Thus the only mechanical energy present at the beginning is the ball’s gravitational energy, E (0) mech = U (0) grav = mg H . As the ball falls down below the height h c = H − L , the cord unfolds and stretches to L = H − h > L , which costs potential energy U cord = k 2 ( H − h − L ) 2 (for h < H − L ) . At first, this elastic energy gain is less than the gravitational energy release due to the ball going down; the difference goes to the ball’s kinetic energy, and the ball continues to accelerate. But later, the elastic energy outgrows the gravitational energy release, the kinetic energy has to decrease and the ball slows down. At the bottom of the ball’s tra- jectory, its speed drops all the way to zero, thus K = 0 and E bot mech = U grav + U cord = mg h bot + k 2 ( H − h bot − L ) 2 . But since the net mechanical energy is con- served, E bot mech = E (0) mech , we must have mgH = mgh bot + k 2 ( H − h bot − L ) 2 ....
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Kaplunovsky - exam 03 - midterm 03 – – Due Apr 4 2007...

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