# Stress / strain

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 Is there something wrong with this thread. Many views but no reply. Is it cause it's such easy question?

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

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