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Energy (Potential) and Oscillation Problem

  • Thread starter yaylee
  • Start date
  • #1
22
0

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!
 

Answers and Replies

  • #2
3,812
92
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.
 
  • #3
22
0
This involved a tedious quadratic equation, but Worked out well, Prana-Arora:
Thank you very much.
 

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