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Visualize a scenario where two springs are laid side by side. The first spring, unstretched, rests at a length of 20cm. The second spring is being stretched to a length of 25cm by a weight, suggesting a force of 50N applied to it - simplify this with a small abstract weight at the end of the spring. Yet another similar weight is depicted beside the second spring, hinting at the possibility of doubling the load to reach 100N. Note there should be an absence of any text or numerical values in the illustration.

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?

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.

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

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

K=F/e

K=50/25-20
K=10N/cm

e=F/K
e=100/10
e=10
Therefore new length =20 + 10 = 30cm

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

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

Absolutely correct!!!

I love the way physics problems are solved,for i love physics,infact.

And : 1.875joules

Lennard Ans: 1.875 joules approximately 1.9 joules

Force=50N ,Extension=25cm-20cm =5cm

So,F=50N/5cm=10N/cm
E=100N/10N/cm=10cm
New length=[original length+E]
=20+10
=30cm

And: 1.875joules