To find the maximum deflection of spring k2 (\(x\)), we can use the principle of conservation of mechanical energy. The potential energy stored in spring k1 when it's compressed is converted into potential energy in spring k2 and kinetic energy of the block when it's launched.
The potential energy stored in a spring is given by:
\[PE = \frac{1}{2}kx^2\]
Where:
\(PE\) is the potential energy stored in the spring.
\(k\) is the spring constant.
\(x\) is the deflection from the equilibrium position.
Initially, spring k1 is compressed by \(0.050 m\) (as given). So, the potential energy stored in it is:
\[PE_{k1} = \frac{1}{2}(2.0 \, \text{N/m})(0.050 \, \text{m})^2 = 0.0025 \, \text{J}\]
This potential energy will be transferred to spring k2 when it's compressed. Therefore:
\[PE_{k1} = PE_{k2} = \frac{1}{2}(1.4 \, \text{N/m})x^2\]
Now, we can solve for \(x\) (the maximum deflection of spring k2):
\[0.0025 \, \text{J} = \frac{1}{2}(1.4 \, \text{N/m})x^2\]
\[0.0025 \, \text{J} = 0.7 \, \text{N/m} \cdot x^2\]
Now, solve for \(x\):
\[x^2 = \frac{0.0025 \, \text{J}}{0.7 \, \text{N/m}}\]
\[x^2 = 0.0035714 \, \text{m}^2\]
\[x = \sqrt{0.0035714 \, \text{m}^2} \approx 0.0598 \, \text{m}\]
So, the maximum deflection of spring k2 (\(x\)) is approximately \(0.0598 \, \text{m}\) or \(5.98 \, \text{cm}\).