Hooke's Law

[Hckyplayer8]

Please see the following. The only reason I stuck this under the Calculus sub is because Webassign suggests that is what is wrong with my answer...even there was no integration done with the example problem.


Please see my work below.

 

[MarkFL]

I would begin by writing:

[MATH]k=\frac{mg}{d}=\frac{(5\text{ kg})\left(9.8\dfrac{\text{m}}{\text{s}^2}\right)}{0.02\text{ m}}=2.45\times10^3\frac{\text{N}}{\text{m}}[/MATH]
(a) [MATH]|F_s|=|mg|[/MATH]
[MATH]2.45\times10^3\frac{\text{N}}{\text{m}}x=(1.5\text{ kg})\left(9.8\dfrac{\text{m}}{\text{s}^2}\right)[/MATH]
[MATH]x=0.006\text{ m}=0.6\text{ cm}\quad\checkmark[/MATH]
(b) [MATH]W=\int_0^{x_1} kx\,dx=\frac{1}{2}kx_1^2[/MATH]
[MATH]W=\frac{1}{2}\cdot2.45\times10^3\frac{\text{N}}{\text{m}}(0.04\text{ m})^2=1.96\text{ J}[/MATH]

 

[Deleted member 4993]

As indicated (in response # 2) above, the displacement for unstretched position is 0. So the lower limit of integration is 0.