Pogo Stick Elastic Energy: Substituting Correct Variables

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SUMMARY

The discussion focuses on calculating the height a 35 kg child will bounce on a pogo stick with a spring constant of 4945 N/m when compressed by 25 cm. The correct approach involves using the conservation of energy principle, equating initial total energy (stored mechanical energy of the spring) to final total energy (gravitational potential energy at the peak height). The final calculated height is 0.45 m, derived from the equation 0.5 * k * s_f^2 = m * g * h, where k is the spring constant, s_f is the compression distance, m is mass, g is acceleration due to gravity, and h is the height.

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Homework Statement


A 35 kg child is jumping on a pogo stick. If the spring has a constant of 4945 N/m and it is compressed 25 cm, how high will the child bounce?

Answer: 0.45 m

Need help substituting the right variable.

Homework Equations



ETi= ETf
Egi + Eki + Eei = Egf + Ekf + Eef
mghi + 0.5mvi2 + 0.5ksi2 = mghf + 0.5mvf2 + 0.5ksf2


The Attempt at a Solution



The real problem I have is setting the defined variables; everything I can solve for. I used trial and error and got 0.45 but I NEED TO KNOW WHY I use the variables (magnitudes in the right place)!

m = 35 kg
k = 4954 N/m
g = 9.8 m/s²
si = 0 m
sf = 0.37 m
hi = 0 m
hf = ?
vi = 0 m/s²
vf = 0 m/s²

I rearranged and solved and I got -0.45 m, some please tell me where what variables I have to substitute correctly! Thank you!
 
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There are a number of different ways to answer this question but from the "relevant equations" you list, you appear to be using "conservation of energy". That's what your first equations says- initial total energy equals final total energy. There are three types of energy involved in such a jump- the gravitational potential energy, kinetic energy, and the stored mechanical energy of the spring. Those are your "E_g", "E_k", and "E_e". The additional "i" and "f" subscripts "initially", at the beginning of the jump, and "finally", at the highest point of the jump. The third equation replaces those with the standard formulas.

We can take the ground as our base for potential energy so the potential energy is 0 there and mgh at h feet at h meters above the ground. At the start and at the highest point, the child is motionless so kinetic energy is 0 at both points. On the ground the stored mechanical energy is, as you say, 0.5ks_f^2. That is, on the ground the total energy is 0+ 0+ 0.5(4945)(.25)^2 and at the highest point of the jump the total energy is (35)(9.88)(h)+ 0+ 0. Set those equal and solve for h.
 
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