Question

AWOOOOOOOO
Motion of an object attached to a spring.

CAUTION:
Work done on a spring versus work done by a spring.
If a spring is stretched from a relaxed position \(x=0\), the force exerted to one end of the spring is in the same direction as the displacement, so the work is positive. In contrast, the work that the spring does on whatever it is attached to is negative.

equilibrium
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\(-x_{max}\)
http://assets.openstudy.com/updates/attachments/5691e7c9e4b0adda82bc178f-nincompoop-1452502888872-screenshot20160111at3.58.51am.png
\(+x_{max} \)
http://assets.openstudy.com/updates/attachments/5691e7c9e4b0adda82bc178f-nincompoop-1452502967757-screenshot20160111at3.59.48am.png

Answer 1

Reviewing the Hooke's Law.

The equilibrium position of the system pertains to the position of an object attached to a spring system when the spring is neither stretched nor compressed. The attached object, a block for an example, is at rest and can be expressed or identified as:

\(x=0 \)

When the spring is disturbed, it typically oscillates and forth and the force it exerts on a block can be modeled mathematically as:

\(F_s=-kx \)

where x is the position of the block relative to its equilibrium position, \(x=0 \), and k is a positive constant called force constant or the spring constant of the spring. So the force required to stretch or compress a spring is proportional to the amount stretch or compression \(x \). This force law for spring is what we refer to as The Hooke's Law.

The value of the \(k \) is a measure the spring's stiffness. So a very stiff spring or spring that is difficult to compress or stretch has a large k value, and for soft spring the k value is small.

A review of algebra. Recall the slope-intercept form \(y = mx+b\), the y becomes the F; the m becomes the k and the x is still x.