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Hooke’s Law for an elastic spring states that the distance a spring stretches varies directly as the force applied. If a force of 180 newtons stretches a spring 5 cm, how much will a force of 270 newtons stretch the same spring?

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Answer:

Simple springs obey Hooke's Law. Hooke's Law states that the force on a spring is proportional to it's displacement from it's resting length:

F = -k*x

Where F is the force, x is how far the string is stretched or compressed, and k is a proportionality constant called the spring constant. The negative sign is because the spring resists stretching.

In this case, we can solve for k using the first case:

(100N) = -k * (5cm)

k = ???

And plug that answer in to find x in the second case:

(360N) = -k * x

    This is a "Hooke's Law" problem: the restoring force required to compress or stretch a spring is proportional to the distance the spring is deformed.  You can figure solve it by setting up a proportion equation and then using the algebraic tool of cross multiplication to solve for the unkown "stretch".

1) set up the proportion equation

100  =  360

5           x

2) cross multiply to get a linear equation

100x = 5*360

3) solve for x

 100x = 1800

       x = 18

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Answer:

The law: F=kx

Therefore,

180N = k * 5cm

k = 180÷5= 36N per cm

Using the known constant with the law,

270=36*x

x= 270 ÷ 36 = 7.5cm

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