Given
Spring constant, k=56 N/m
The weight of the block, F=18 N
To find
The stretch in the spring
Explanation
By hook's law,
[tex]F=kx[/tex]where x is amount the spring is stretched.
Putting the values,
[tex]\begin{gathered} 18=56x \\ \Rightarrow x=0.32 \end{gathered}[/tex]Conclusion
The amount by which the spring is stretched is 0.32 m
