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ActivPhysics OnLine Problem 5.4: Inverse Bungee Jumper

ActivPhysics OnLine Problem 5.4: Inverse Bungee Jumper Choose xy coordinates in the usual way. y x upwards : + y downwards: - y In this problem we are assuming there is no friction. Problem 5.4: Question 1 Do the simulation.

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ActivPhysics OnLine Problem 5.4: Inverse Bungee Jumper

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  1. ActivPhysics OnLine Problem 5.4: Inverse Bungee Jumper Choose xy coordinates in the usual way. y x upwards : + y downwards: - y In this problem we are assuming there is no friction.

  2. Problem 5.4: Question 1 Do the simulation. Notice : (a) the woman’s gravitational potential energy increases as her height above the floor increases; (b) the spring’s potential energy increases as the spring is stretched; (c) the woman’s kinetic energy is maximized when she is near the middle of an oscillation.

  3. Problem 5.4: Question 2 Construct an energy equation for the woman’s motion. What is the force constant of the spring, k, if her head just touches the overhead support without injury? The system’s total energy E is the sum of kinetic energy K, grav. potential energy Ug, and the spring’s potential energy Us. E = K + Ug + Us

  4. (This problem is similar to Randall Knight’s Physics for Scientists and Engineers, Example 10.7 on pages 285-286.) E = K + Ug + Us Let ye = the equilibrium position of the spring, i.e. the position at which there is no restoring force. E = m v2/2+ mgy + k (y - ye)2 /2

  5. Since total energy is conserved, the energy of the system when she is at the top position y2 = energy of the system when she is at the bottom position y1. mv22/2+ mgy2 + k (y2 - ye)2/2 = mv12/2+ mgy1 +k (y1 -ye)2/2 Her kinetic energy should be 0 at the top and is 0 also at the bottom where she is at rest. mgy2 + k (y2 - ye)2/2 = mgy1 +k (y1 -ye)2/2

  6. The gravitational potential energy also must be 0 at the bottom where y = 0. mgy2 + k (y2 - ye)2/2 = k (y1 -ye)2/2 k (y1 -ye)2/2 - k (y2 - ye)2/2 = mgy2 k [(y1 -ye)2/2 - (y2 - ye)2/2 ] = mgy2 Substitute in values. m = 50 kg g = 10 m/s2

  7. Since the y values are zero at the floor y2 = 4.0 m at the top y1 - ye = - 3.5 m when the spring is stretched and her feet touch the floor. At the top the spring is slightly compressed and y2 - ye = + 0.5 m. k [(-3.5m)2 +(0.5m)2 ] = (2)(50 kg)(4.0m)(10m/s2) k= 4000 N/ 12 m2

  8. k= 333 N/ m The simulation only proceeds in steps of 5.0 N/m. Notice on the simulation if k= 330 N/ m, her head does not quite reach the top, but if k= 335 N/ m, then she hits her head and is injured.

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