Calculating Potential & Kinetic Energies of a 80kg Man Swinging on a Swing

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Homework Help Overview

The discussion revolves around calculating the potential and kinetic energies of an 80 kg man swinging on a swing with a 1 m amplitude. The problem specifies that he completes 15 swings in 60 seconds, and participants are tasked with finding the energies after 1/12 of a period.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the maximum energy calculated as 100 J and express confusion regarding the potential and kinetic energy values derived from their calculations. There are attempts to clarify the equations and the relationships between variables such as time, amplitude, and energy.

Discussion Status

Some participants have provided guidance on showing full calculations to identify errors, while others suggest simplifying the approach to finding velocity and energy. Multiple interpretations of the problem setup and calculations are being explored, but no consensus has been reached.

Contextual Notes

Participants note potential complications in their calculations and question the assumptions made regarding the swing's length and the relationship between distance and time. There is an acknowledgment of the timing of posts affecting response rates.

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Homework Statement


80kg mass man is swinging on a swing. His amplitude is 1m. in 60 seconds he does 15 swings. Find the potential and kinetic energy after 1/12 period.


Homework Equations

W(p)=mgh or kx^2/2 W(k)=mv^2/2 W(max)=Fr^2mg/2l



The Attempt at a Solution

i Found out that maximum energy is 100J. But when i am calculating energy of petencial or kinetic i get 1/3 or 2/3 of maximum...or something even more horrid. =]
 
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oh c'mon! there must be someone out there knowing how to do this...
 
Hi Patrickas! :smile:

(try using the X2 tag just above the Reply box :wink:)
Patrickas said:

Homework Statement


80kg mass man is swinging on a swing. His amplitude is 1m. in 60 seconds he does 15 swings. Find the potential and kinetic energy after 1/12 period.


Homework Equations

W(p)=mgh or kx^2/2 W(k)=mv^2/2 W(max)=Fr^2mg/2l



The Attempt at a Solution

i Found out that maximum energy is 100J. But when i am calculating energy of petencial or kinetic i get 1/3 or 2/3 of maximum...or something even more horrid. =]

Patrickas said:
oh c'mon! there must be someone out there knowing how to do this...

erm :redface: … you can't post around 2am New York time and expect an answer 4 hours later! :rolleyes:

anwyay, show us your full calculations, and then we'll see what went wrong, and we'll know how to help. :smile:
 
tiny-tim said:
Hi Patrickas! :smile:

(try using the X2 tag just above the Reply box :wink:)




erm :redface: … you can't post around 2am New York time and expect an answer 4 hours later! :rolleyes:

anwyay, show us your full calculations, and then we'll see what went wrong, and we'll know how to help. :smile:



yea, but there are a lot more people out there not from new york, who might know.
 
hmm. Anyone..?
 
Show us your full calculations, and then we'll see what went wrong, and we'll know how to help. :smile:
 
T=60/15=4s
t=4/12=1/3s

So i know that W=kx^2/2 Well then i tried getting k, k=mg/l , now i needed the lengh of swings so i figured mv^2/r=mgr/l l=r^2g/v^2, now i need speed =D. v=s/t is s= 4*r? , or s=2*3.14*r? and t=1/3. Anyways i sense that something here is wrong. The numbers don't add up properly.

Assuming that s=2*3.14*r i can get T=2*3.14*\sqrt{\frac{l}{g}} and get that l(t=4s)=4m Then i get that maximum energy is 100J. Now what...

Ahh... i used to calculate s=4r and not 2*3.14*r. I think it works out.

W=\frac{mgt2*3.14r}{l2T}

When i put t=1/3 i get that potencial or kinetic(don't know which..) is 25J ! Which means that other is 75J Now which is which...?
 
Last edited:
tiny-tim said:
Show us your full calculations, and then we'll see what went wrong, and we'll know how to help. :smile:

Hey thanks for the help! I have been trying to solve this one for days! hehe
 
Hi Patrickas! :smile:

You're making this very complicated. :redface:

You know T = 4, A = 1, and so distance x = 1*cos(2πt/4).

So you can find v … then KE = 1/2 mv2, and PE = … ? :smile:
 
  • #10
tiny-tim said:
Hi Patrickas! :smile:

You're making this very complicated. :redface:

You know T = 4, A = 1, and so distance x = 1*cos(2πt/4).

So you can find v … then KE = 1/2 mv2, and PE = … ? :smile:


O yea! Thanks! It should be much faster and easier to solve those problems now.
 

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