A spring has a force constant of 525 N/m. A force of 375 N is applied to the spring. Find the elongation of the spring and its elastic potential energy. By how much will the spring elongate when stretched by force of 350 N? What is its elastic energy?

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I suppose that the spring follows Hooke's Law:

`F=kx,`

where `F` is the force (it is given), `k` is the spring force constant (given also) and `x` is the length of the elongation or compression. Using this formula we can find the length of an elongation,

`x=F/k.`

Also it is known that for such a springs the elastic potential energy is equal to

`(kx^2)/2=(F^2)/(2k).`

 

Now perform the calculations.

For the first case, the elongation `x` is `F/k=375/525 approx 0.71(m).`
The elastic potential energy is `(375^2)/(2*525) approx 134 (J).`

For the second case, the elongation x is `F/k=350/525 approx 0.67(m).`
The elastic potential energy is `(350^2)/(2*525) approx 117 (J).`

There is no surprise that for the less force the elongation occurred to be less also, and the same is observed for the elastic potential energy.

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