Stress / strain

  • Thread starter lemon
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  • #1
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1. A mass of 100g is suspended from a light spring and extends the spring by 4cm when the mass in in equilibrium.
a) Find the spring constant k of the spring?
b) How much elastic energy is now stored in the spring?
c) What would be the extension of the spring if a mass of 170g were suspended?
d) How much extra energy would now be stored in the spring?
e) Why is more work needed to stretch a spring by 1cm from its unextended length, than is required to stretch it by an extra 1cm when the spring is already extended?



2. Let deltaX = e
F=ke
Ep=1/2Fe
Stress=F/A
Strain=e/L
E=stress/strain
F=ma

100g = 0.1kg
4cm = 0.04m
Let gravity = -10m/s



3. a) F=0.1 x 10 = 1N
k=10N/0.04m = 250N/m

b) E=1/2 x 10N x 0.04m = 0.2J
c) F=0.170 x 10 = 1.7N
e = 1.7N/250N/m = 0.006.8m
d) E = 1/2 x 10N x 0.170m = 0.85J
e) I know this holds true from my experience in reality, but when I consider the force extension graph and hook's law, it tells me that up until the elastic limit the same force is required to achieve the same extension. But considering E=1/2 x stretching force x extension, from this equation it is clear that an increase in extension will increase elastic potential energy.


Could someone kindly check and guide, please?
Thank you :bugeye:
 

Answers and Replies

  • #2
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Is there something wrong with this thread. Many views but no reply. Is it cause it's such easy question?
 
  • #3
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Ok. You'll better ignore the above post. I guess I really was under too much stress and strain.
So, hopefully to put it right:

mass=0.10kg
Δx=0.04m

a) Find the spring constant k of the spring?
F=ma
0.10x10=1.0N

F=kΔx
k=F/Δx
1.0/0.04=25N/m
 
  • #4
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b) How much elastic energy is now stored in the spring?
Ep=1/2FΔx
1/21.0x0.04=0.02J

c) What would be the extension of the spring if a mass of 170g were suspended?
F=ma
0.170kgx10=1.70N
Δx=F/k
1.70/25=0.068m

d) How much extra energy would now be stored in the spring?
Ep=1/2FΔx
1/2x1.70x0.68=0.578m
 

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