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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

E

Stress=F/A

Strain=e/L

E=stress/strain

F=ma

F=ke

E

_{p}=1/2FeStress=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.

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