(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)
Answer:
[tex]d=\frac{2g(m1-m2)}{k} [/tex]
Explanation:
For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:
[tex]E_{i}=E_{f} [/tex]
[tex] K_{i}+U_{i}=K_{f}+U_{f} [/tex] (1)
With [tex]K_{i} [/tex] the initial kinetic energy, [tex] U_{i}[/tex] the initial potential energy, [tex]K_{f} [/tex] the final kinetic energy and [tex]U_{f} [/tex] the final potential energy. Note that initialy the masses are at rest so [tex]K_{i} = 0 [/tex], when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so [tex]K_{f} =0 [/tex]. So, equation (1) becomes:
[tex]U_{i}=U_{f} [/tex] (2)
At initial moment all the potential energy is gravitational because the spring is not stretched so [tex]U_{i}=U_{gi} [/tex] and at final moment we have potential gravitational energy and potential elastic energy so [tex] U_{f}=U_{gf}+U_{ef}[/tex], using this on (2)
[tex] U_{gi}=U_{gf}+U_{ef}[/tex] (3)
Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), [tex] U_{gi}=0[/tex] and we have by (3) :
[tex]0= U_{gf}+U_{ef} [/tex] (4)
Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:
[tex]0=-m1gd+m2gd+\frac{kd^{2}}{2} [/tex]
dividing both sides by d
[tex]0=-m1g+m2g+\frac{kd}{2}[/tex]
[tex]g(m1-m2)= \frac{kd}{2}[/tex]
[tex]d=\frac{2g(m1-m2)}{k} [/tex], with k the constant of the spring and g the gravitational acceleration.
