Calculating Spring Compression: Finding Height and Max Velocity

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The discussion revolves around calculating the height a 16 lb weight will reach when a spring is compressed by 6 inches, using conservation of energy principles. Participants emphasize the importance of considering both potential and kinetic energy in the calculations, noting that the spring's potential energy must equal the gravitational potential energy at the peak height. There is confusion regarding the units of the spring constant and the need for conversions between US and SI units. Ultimately, one participant claims to have solved the problem independently, despite the ongoing debate about unit correctness. The conversation highlights the complexities of energy conservation in spring mechanics.
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The setup shown is at rest. If the spring is compressed 6in. how high (h) above Equi. will the 16lb. weight reach? Max Velocity?

I'm using U_1,2=.5k(x1)^2-.5k(x2)^2 (Potential Energy)

I know it is a simple problem but there is something i am not catching. so far I have tried this:

16lb.(h+6)+[.5(15)1.06in.-.5(15)6in.]=0
1.06in=distance weight alone compresses spring
I know this is WRONG. Where am I WRONG?
 
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Conservation of energy is your answer :

kx²/2 is the potential energy when a spring is compressed over a distance x. Then, the spring 'fires off' the object. Ofcourse this object will fly upwards but gravity will eventually stop this motion. So, the initial energy will be needed in order to 'fight' against gravity. In the conservation law, you will need to evaluate both kinetic and potential ebergy at the biginning and at the end of the motion, so you will also need to bring in gravity. Do you know the gravitational potential energy ? If so, your work is done...


marlon
 
T1+V1=T2+V2
T1=.5mv^2=.5(.497)0=0
T2=.5(.497)*?
V1=mgh=.497(32.2)0=0
V2=.497(32.2)h

This is what you are saying right?
If so what goes in the "?" spot?
If not, Why not?
 
No, you are wrong.

beginning : the spring is compressed 6 inches and the kinetic energy is ZERO

end : the object is a distance h above the equilibrium point and the kinetic energy is zero.

So you have : kx²/2 = mgh and solve this for h

marlon
 
(15lb/in.*6in)/2=16lb*h
h=16.87in
Actual answer is 45.2in

We are both missing something.
 
are the units of the spring constant Newton/meter ? because if so, you will need to convert your units into meters and kilograms...

marlon
 
He needn't,marlon,we do...:-p Units are okay fixed,so the calculations should go easily.Maybe if he translated from US to SI we'd be able to follow his #-s


Daniel.
 
dextercioby said:
He needn't,marlon,we do...:-p Units are okay fixed,so the calculations should go easily.Maybe if he translated from US to SI we'd be able to follow his #-s


Daniel.

Indeed, :smile:

However, isn't the spring constant given in #/m

what the hell is this # ? :smile:

marlon
 
You guys going to help or not? :mad:
 
Last edited:
  • #10
If I'm not mistaking,they use it as pounce avoirdupoids per inch.They should use a unit for force,but they use a unit for mass.I think that's incorrect,or even dumb.

Daniel.
 
  • #11
k=2626.8N/m
x=.1524m
weight collar=71.04N
g=9.81m/s^2
 
  • #12
VSCCEGR said:
k=2626.8N/m
x=.1524m
weight collar=71.04N

Look; the solution is definitely the way i presented it to you. There must be something wrong with the units.

marlon
 
  • #13
Never Mind. I Solved It With Out You All.
 
  • #14
VSCCEGR said:
Never Mind. I Solved It With Out You All.

yeah, whatever, :rolleyes:
 
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