Answer:
The spring constant is approximately [tex]681\; {\rm N \cdot m^{-1}}[/tex].
The spring would stretch approximately [tex]1.44\; {\rm cm}[/tex].
Approximately [tex]2.76\; {\rm J}[/tex] of energy is required to stretch this spring to the required position.
(Assumption: [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Make sure all displacements are measured in standard units:
[tex]2.88\; {\rm cm} = 0.0288\; {\rm m}[/tex].
[tex]9.00\; {\rm cm} = 0.0900\; {\rm m}[/tex].
To find the spring constant [tex]k[/tex], divide force on the spring [tex]F[/tex] by displacement [tex]x[/tex]. That is: [tex]k = F / x[/tex].
In this question, force on the spring is equal to the weight of the object. To find the weight of the object, multiply mass [tex]m[/tex] by [tex]g[/tex]:
[tex]\begin{aligned}(\text{weight}) &= m\, g \\ &= (2.00\; {\rm kg}) \, (9.81\; {\rm N\cdot kg^{-1}}) \\ &= 19.62\; {\rm N}\end{aligned}[/tex].
Substitute [tex]F = (\text{weight}) = 19.62\; {\rm N}[/tex] into the equation [tex]k = F / x[/tex] to find the spring constant [tex]k[/tex]:
[tex]\begin{aligned} k &= \frac{F}{x} \\ &= \frac{19.62\; {\rm N}}{0.0288\; {\rm m}} = 681.25\; {\rm N\cdot m^{-1}}\end{aligned}[/tex].
By Hooke's Law, as long as spring constant [tex]k[/tex] is fixed, displacement of the spring would be proportional to the external force on the spring.
If [tex]g[/tex] is fixed, the weight of an object will be proportional to its mass. Hence, the weight of the [tex]1.00\; {\rm kg}[/tex] object would be [tex](1/2)[/tex] that of the [tex]2.00\; {\rm kg}[/tex] object.
Therefore, reducing the mass of the object on the spring by [tex](1/2)[/tex] would also reduce the displacement of the spring by [tex](1/2)\![/tex].
If the spring constant [tex]k[/tex] is fixed, the work required to stretch the spring by [tex]x[/tex] is given by the formula [tex](1/2)\, k\, x^{2}[/tex].
In this example, it is already found that [tex]k = 681.25\; {\rm N \cdot m^{-1}}[/tex]. The energy required to stretch the spring by [tex]x = 0.0900\; {\rm m}[/tex] would be:
[tex]\begin{aligned} \frac{1}{2}\, k\, x^{2} =\; & \frac{1}{2}\, (681.25\; {\rm N \cdot m^{-1}})\, (0.0900\; {\rm m})^{2} \\ \approx \; & 2.76\; {\rm J}\end{aligned}[/tex].
