A spring exerts a force of 6 N when stretched 3 m beyond its natural length. How much work is required to stretch the spring 2 m beyond its natural length?

4Joules

Step-by-step explanation:

According to Hooke's law which states that extension of an elastic material is directly proportional to the applied force provide that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

K is the elastic constant

e is the extension

If a spring exerts a force of 6 N when stretched 3 m beyond its natural length, its elastic constant 'k'

can be gotten using k = f/e where

F = 6N, e = 3m

K = 6N/3m

K = 2N/m

Work done on an elastic string is calculated using 1/2ke².

If the spring is stretched 2 m beyond its natural length, the work done on the spring will be;

1/2× 2× (2)²

= 4Joules

the answer is b.

64

step-by-step explanation:

c=2πr=2·π·8≈64

ιт ιѕ 6 6/23

step-by-step explanation:

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