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...

Hello!

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...

Hello!

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.