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Integrals and Work: Example 2 - Hooke's Law

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Published: 09 Jan 2020 โ€บ Updated: 09 Jan 2020Integrals and Work: Example 2 - Hooke's Law

Integrals and Work: Example 2 - Hooke's Law

In this video I go over another example on work and this time deal with Hooke's Law. Hooke's Law states that the force required to maintain a spring stretched x units beyond its natural length is proportional to x and can be written as f(x) = kx where k is a positive constant and is known as the spring constant. The example that I cover is an example involving stretching a spring and determining the amount of work that was required.


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Integrals and Work: Example 2 - Hooke's Law

Integrals and Work Example 2.jpeg

The following example requires using a law from physics called Hooke's Law which states that the force required to maintain a spring stretched x units beyond its natural length is proportional to x:

f(x) = kx

where k is a positive constant (called the spring constant).

Hooke's Law holds provided that x is not too large.

(a) Natural position of the spring

(b) Stretched position of spring

Example:

A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How much work is done in stretching the spring from 15 cm to 18 cm?

Solution:

Leave Integrals and Work: Example 2 - Hooke's Law to:

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