Energy and simple harmonic motion

In summary: L0)*k In summary, the jumper is 10 meters from the water when he reaches the lowest point in his fall.
  • #1
jacksonpeeble
Gold Member
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2

Homework Statement


A 74.4-kg bungee jumper is standing on a tall platform (h0 = 46.6 m). The bungee cord has an unstrained length of L0 = 9.54 m and, when stretched, behaves like an ideal spring with a spring constant of k = 61.8 N/m. The jumper falls from rest, and it is assumed that the only forces acting on him are his weight and, for the latter part of the descent, the elastic force of the bungee cord. Determine how far the bungee jumper is from the water when he reaches the lowest point in his fall.



Homework Equations


mghf+.5ky2=mgh0


The Attempt at a Solution


No idea, made several attempts with results close to 10.
 
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  • #2
Try finding how much the cord extends when the person jumps.
 
  • #3
How far is the jumper from his initial position when he is at the end of the maximally stretched cord?

ehild
 
  • #4
in my opinion you can divide the jump into 2 stages: 1) free falling (H of free falling = L= 9.54) 2) movement on y asis and in this stage you can set equation (using Newton's laws) weight force (m*g) - elastic force = m*a
 
  • #5
Thank you everyone for your advice - I am curious - in what order should I go about doing these operations.

For instance (without necessarily plugging in numbers for me, which I know would defeat the purpose of me actually learning the material), is there a step-by-step procedure that should be followed (and other equations)?
 
  • #6
in the first stage you can solve nothing but in second one you should use formula F(elastic) = k*(L-L0) and inferring from Newton's second law, that is any force = m*a we can get equation k*(L-L0) = m*a the acceleration is g=10m/s^2 because it is a vertical falling (i am not sure about my opinion but i think that..)
transforming the equation we can get that L = (m*g+k*L0)*k and a length between the water and the reachest point is H0-L
 

Related to Energy and simple harmonic motion

1. What is energy and how does it relate to simple harmonic motion?

Energy is the ability to do work or cause change. In simple harmonic motion, energy is constantly being exchanged between potential energy and kinetic energy as an object oscillates back and forth.

2. How is potential energy involved in simple harmonic motion?

Potential energy is stored energy that an object possesses due to its position or configuration. In simple harmonic motion, potential energy is at its maximum when the object is at its furthest displacement from equilibrium, and decreases as the object approaches equilibrium.

3. What is the formula for calculating the period of simple harmonic motion?

The formula for calculating the period (T) of simple harmonic motion is T = 2π√(m/k), where m is the mass of the object and k is the spring constant.

4. Can energy be lost in simple harmonic motion?

In an ideal system, energy is conserved and cannot be lost. However, in real systems, energy may be lost due to friction or other external forces, causing the amplitude of the oscillation to decrease over time.

5. How does the amplitude of simple harmonic motion affect the energy of the system?

The amplitude of simple harmonic motion is directly proportional to the energy of the system. As the amplitude increases, so does the energy, and vice versa. This relationship can be seen in the equation for potential energy: PE = ½kx², where x is the amplitude.

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