Respuesta :
Answer:
x₂=0.44m
Explanation:
First, we calculate the length the spring is stretch when the first block is hung from it:
[tex]\Delta x_1=0.50m-0.35m=0.15m[/tex]
Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:
[tex]k\Delta x_1=m_1g[/tex]
Solving for the spring constant k, we get:
[tex]k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}[/tex]
Next, we use the same relationship, but for the second block, to find the value of the stretched length:
[tex]k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m[/tex]
Finally, we sum this to the unstretched length to obtain the length of the spring:
[tex]x_2=0.35m+0.088m=0.44m[/tex]
In words, the length of the spring when the second block is hung from it, is 0.44m.
