Calculating Strain Energy in a Hanging Spring

In summary, a spring stretches 2.6 cm before coming to a temporary stop and the strain energy is half its maximum value at that point.
  • #1
Nemo's
69
0

Homework Statement


A spring hangs vertically from a fixed point. When a mass of 94g is suspended from the spring the spring stretches by 2.6 cm before temporarily coming to rest. Determine the extension of the spring at which the strain energy is half its maximum value.


Homework Equations



F=0.5kx^2

The Attempt at a Solution


Loss in Gravitational Potential Energy = Gain in Strain Energy
So Half Strain energy = Half Gravitational Potential Energy
0.5k(.026*10^-2)^2=0.012 so k=6000/169
and since k=F/x using F=0.094*9.8 to get x
 
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  • #2
Nemo's said:
0.5k(.026*10^-2)^2=0.012 so k=6000/169
I don't understand your working there. Pls post full details. .026*10^-2 doesn't look right - do you mean .026? For k, I get about 70N/m.
 
  • #3
Oops sorry. Yes I meant 0.026 but my value for k is 6000/169 (approx. 35.5)
Then I used the Weight force and my k in F=kx (hook's law) to get x so (0.094*9.8)/(35.5)=x
 
  • #4
Nemo's said:
Oops sorry. Yes I meant 0.026 but my value for k is 6000/169 (approx. 35.5)
OK, but I asked you to explain how you got that equation. Where does the .012 come from? That looks to me like half the strain energy at max extension, but to compute k you need to be using the whole strain energy at max extension.
 
  • #5
edited post
 
Last edited:
  • #6
haruspex said:
Where does the .012 come from? That looks to me like half the strain energy at max extension
Yes the .012 is half the gravitational potential energy lost (mgh) which is equal to half the strain energy gained(1/2 k x^2) (Using conservation of energy)
Nemo's said:
Loss in Gravitational Potential Energy = Gain in Strain Energy
So Half Strain energy = Half Gravitational Potential Energy
0.5k(.026)^2=0.012
haruspex said:
but to compute k you need to be using the whole strain energy at max extension.
So I wrongly equated the whole strain energy to half the strain energy. I corrected it now:
.25k(.026)^2=.012
Now I get k=71
haruspex said:
For k, I get about 70N/m.
Is that how you got it ?
 
  • #7
Nemo's said:
.25k(.026)^2=.012
Now I get k=71

Is that how you got it ?
No. You are told:
A spring hangs vertically from a fixed point. When a mass of 94g is suspended from the spring the spring stretches by 2.6 cm before temporarily coming to rest.
That is enough information to calculate k. No mention is made of any half amounts of energy yet.
 
  • #8
O.K but in the beginning I tried using F=ke to get k. for F= .094*9.8 and e=.026, k=35.4
Why didn't I get the true value for k ?
 
  • #9
Nemo's said:
O.K but in the beginning I tried using F=ke to get k. for F= .094*9.8 and e=.026, k=35.4
Why didn't I get the true value for k ?
Because that equation assumes the extension is to the equilibrium point (zero acceleration), but the extension given is to zero KE point.
 
  • #10
Thank you so much :D
 

Related to Calculating Strain Energy in a Hanging Spring

1. What is strain energy in a spring?

Strain energy in a spring is the potential energy stored in a spring when it is deformed or stretched from its original position. It is also known as elastic potential energy.

2. How is strain energy calculated in a spring?

The strain energy in a spring can be calculated using the formula U = (1/2)kx^2, where U is the strain energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

3. What factors affect the strain energy in a spring?

The strain energy in a spring is affected by the spring constant, the displacement of the spring, and the material properties of the spring such as its stiffness and elasticity.

4. How does strain energy relate to Hooke's Law?

Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. This means that as the spring is stretched or compressed, strain energy is stored in the spring and can be released when the spring returns to its original position.

5. What are some real-life applications of strain energy in a spring?

Strain energy in a spring is used in various applications, such as in shock absorbers, car suspensions, and door hinges. It is also used in mechanical devices like watches and clocks to store energy and release it in a controlled manner.

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