Work, Energy, and the CWE Theorem

AI Thread Summary
The discussion focuses on a bungee jumping scenario involving a student of mass m dropping from a height h with a bungee cord characterized by a spring constant k = 2mg/h. The initial total mechanical energy is determined to be solely potential energy since the student starts at rest. Conservation of energy principles indicate that the student arrives at the ground with zero kinetic energy, as all potential energy converts to kinetic energy at the lowest point. The maximum speed occurs at the lowest point of the jump, where potential energy is minimized. It is clarified that the safety of the bungee cord for different masses depends on the spring constant and the resulting acceleration, not just the mass itself.
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A student of mass m drops off a tower of height h, tethered to the bungee cord. The elasticity of the bungee cord is hosen so that its effective spring constant is k = 2mg/h. The jump is begun a zero speed at altitude h with the bungee cord fastened vertically above her, barely taut(with no slack), but under no initial tension.

(a) Choose a convenient coordinate systerm to analyze the problem.

- | h k = 2mg/h kx = F(x). I'm not sure what they are
- | want here because to me we
- | put it all in terms of x
- | as it's the only direction
- | considered.
- |
- O


(b) What is the initial total mechanical energy of the student?
I used the Energy for a simple harmonic oscillator and get..

E = 1/2(k)(x)^2
E = (mgx^2)/2

(c) SHow that the student will arrive at the ground with zero kinetic energy.
... I'm not sure about this one

(d) At what height above the ground is the speed of the student a maximum. I think it's at the point when it changes direction and oscillates back

(e) Will the same bungee cord be safe to use for a jump from the same tower(of height h) for a person with mass m' < m? With mass m' > m? Explain your reasoning.
No for both. The force of the spring doesn't depend on the mass, but only on the displacement and spring constant of the spring.

Is the work correct, not exactly, or absolutely wrong =\
 
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(a) If you denote the student's altitude by x, then they want to know when x is zero. Could be when the student is at the top or bottom. They also want to know whether you count forces as positive when pointing up or down.

(b) Since the student is not moving at the beginning, initial total energy = initial potential energy.

(c) You can use conservation of energy. Zero kinetic energy means 'no motion'. See?

(d) I think your answer is wrong. Because when he changes direction, the speed is zero.

(e) I think your answer is wrong. They do not ask for the force in the cord, but whether he hits the ground. Better use the info given in (c)...
 


Overall, the response provided seems to be on the right track and shows a good understanding of the concepts involved in the problem. However, there are a few areas that could use some clarification or improvement:

a) The chosen coordinate system seems to be fine, but it would be helpful to label the axes and clearly define what each variable represents. For example, x could represent the displacement from the initial position, while h could represent the height above the ground.

b) The initial total mechanical energy should include both potential and kinetic energy. Since the student starts at rest, the initial kinetic energy should be zero and the total mechanical energy should only be the potential energy at the initial height h.

c) To show that the student arrives at the ground with zero kinetic energy, you can use the conservation of mechanical energy principle. This means that the initial total mechanical energy is equal to the final total mechanical energy (which is just the potential energy at the final height, since the kinetic energy is zero). Therefore, at the ground, the kinetic energy must be zero.

d) The maximum speed of the student will occur at the point when the potential energy is at its minimum and all of the potential energy has been converted to kinetic energy. This occurs when the student is at the lowest point of the oscillation, which is at the ground.

e) The reasoning provided for both cases is incorrect. The spring constant does not depend on the mass, but the acceleration of the student does. A lighter person will experience a higher acceleration and therefore may require a different spring constant to ensure a safe and comfortable jump. Similarly, a heavier person may require a higher spring constant to prevent them from hitting the ground too hard.
 
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