The sum of the energy in a isolated system is always constant, according to the Law of Conservation of Energy
[tex]An \ expression \ for \ \mathbf{m} \ is \ m= \mathbf{\dfrac{ k \cdot d^2}{\left(2\cdot g \cdot (h + d) +v^2 \right) }}[/tex]
Reason:
According to the Law of Conservation of Energy, we have;
Mechanical energy of ball = Elastic potential energy gained by spring
M.E. = P.E. + K.E. = [tex]P.E._{spring}[/tex]
The initial potential energy of the ball relative to the compressed spring, P.E. is given as follows;
P.E. = m·g·(h + d)
[tex]The \ kinetic \ energy \ of \ the \ ball\ K.E. = \dfrac{1}{2} \cdot m \cdot v^2[/tex]
[tex]Potential \ energy \ given \ to \ the \ spring \ P.E._{spring} = \dfrac{1}{2} \cdot k \cdot d^2[/tex]
Therefore, we get;
[tex]M.E. = m\cdot g \cdot (h + d) + \dfrac{1}{2} \cdot m \cdot v^2 = \dfrac{1}{2} \cdot k \cdot d^2[/tex]
Which gives;
[tex]m\cdot \left(g \cdot (h + d) + \dfrac{1}{2} \cdot v^2 \right) = \dfrac{1}{2} \cdot k \cdot d^2[/tex]
[tex]m=\dfrac{\dfrac{1}{2} \cdot k \cdot d^2}{\left(g \cdot (h + d) + \dfrac{1}{2} \cdot v^2 \right) } = \dfrac{ k \cdot d^2}{\left(2\cdot g \cdot (h + d) +v^2 \right) }[/tex]
The expression for m in terms of g, h, d, k, and v is therefore;
[tex]m= \dfrac{ k \cdot d^2}{\left(2\cdot g \cdot (h + d) +v^2 \right) }[/tex]
Learn more about the Law of Conservation of Energy here:
https://brainly.com/question/24298627
