# Physic

A spring 20cm long is stretched to 25cm by a load of 50N.What will be its length when stretched by 100N assuming that the elastic limit is not reached?

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1. k = 50N/(25-20)cm = 10N/cm.

L = 20cm + (100N/10N) * 1cm = 30 cm.

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posted by Henry
2. A spring of force constant 1500newton per metre is acted by a constant force of 75newton.calculate the potential enery stored in the spring.

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posted by Philip
3. 10cm

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4. Force = 50N, Extension = 25cm - 20cm = 5cm
so, F= 50N/5cm = 10N/cm

E=100N/10N/cm = 10cm
new length = ( original length + E)
= 20cm + 10cm
=30cm.

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posted by Prevail
5. And: 1.875joules

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6. And : 1.875joules

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7. Lennard Ans: 1.875 joules approximately 1.9 joules

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8. E=25-20=5.so,f=50N/5cm=10N/cm new length =(original length +E)=20cm+10cm=30cm

E=(25cm-20cm)=5cm so,f=50N/5cm=10N/cm E=100N/10N/cm=10cm new length=(original
length+E)=20cm+10cm=30cm

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9. 0.05m

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