Energy (Potential) and Oscillation Problem

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SUMMARY

The problem involves a clown of mass 76.8 kg jumping from a height of 17.9 m onto a net with a spring constant of 16,900 N/m. The initial potential energy (U) is calculated using U = mgh, while the kinetic energy (KE) just before impact is given by KE = (1/2)mv^2. The correct approach to determine the maximum stretch (A) of the net requires adjusting the energy conservation equation to account for the potential energy change when the clown comes to rest on the net, leading to the equation mgh = (1/2)kA^2 + mgA, which simplifies to a quadratic equation for A.

PREREQUISITES
  • Understanding of gravitational potential energy (U = mgh)
  • Knowledge of kinetic energy (KE = (1/2)mv^2)
  • Familiarity with spring mechanics and Hooke's Law (E = (1/2)kA^2)
  • Ability to solve quadratic equations
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems
  • Learn about the dynamics of simple harmonic motion and spring systems
  • Explore the derivation and application of quadratic equations in physics problems
  • Investigate the effects of air resistance on falling objects and energy loss
USEFUL FOR

Students in physics, particularly those studying mechanics and energy conservation, as well as educators looking for practical examples of energy transfer in spring systems.

yaylee
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Homework Statement


Problem: A clown of mass M = 76.8 kg jumps off a tower at height H = 17.9 m above a net that is stretched horizontally. The net acts like a spring with spring constant k = 16900 N/m. How far will the net stretch before the clown comes instantaneously to rest?

Homework Equations


U = mgh
KE = (1/2)mv^2
Energy (Total) of a Spring (Under Simple Harmonic Motion) = (1/2)kA^2, where A = amplitude in meters.


The Attempt at a Solution


Initially: the clown possesses U = mgh Joules.
After: Right before falling down on the net, he possesses U = KE = (1/2)mv^2 Joules, by Conservation of Energy. (We will call this point (arbitrary), U = O.
At the point where the net stretches:
E = 1/2(k)A^2

Since energy is conserved (none lost to air friction, etc.), mgh = (1/2)kA^2. Solving for A, I get 1.2 meters. (which was marked incorrect)

If there is any hint/tips, I would greatly appreciate it!
 
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H is the height of the clown from the ground. When the clown falls on the net and comes instantaneously at rest (the condition for maximum stretching of net), the clown still posses potential energy if U=0 at the ground. A better choice would be to select U=0 at the net and add a term -mgA to the RHS of your equation mgh = (1/2)kA^2.
 
This involved a tedious quadratic equation, but Worked out well, Prana-Arora:
Thank you very much.
 

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