Answer :
A spring with a constant k stores potential energy when it is compressed a distance x. That potential energy is given by the formula:
[tex]U=\frac{1}{2}kx^2[/tex]On the other hand, that potential energy comes from the kinetic energy of the block. If the block has a mass m and an initial speed v, then its kinetic energy is equal to:
[tex]K=\frac{1}{2}mv^2[/tex]Substitute k=955 N/m and x=0.10 m to find the potential energy stored in the spring after the collision:
[tex]\begin{gathered} U=\frac{1}{2}(955\frac{N}{m})(0.10m)^2 \\ =\frac{1}{2}(955\frac{N}{m})(0.01m^2) \\ =\frac{1}{2}\cdot9.55N\cdot m \\ =4.775J \end{gathered}[/tex]Isolate v from the equation for the kinetic energy of the block:
[tex]v=\sqrt[]{\frac{2K}{m}}[/tex]Since all the potential energy of the spring corresponds to the initial kinetic energy of the block, substitute K=4.775J and m=1.70kg to find the initial speed of the block:
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot4.775J}{1.70\operatorname{kg}}} \\ =\sqrt[]{\frac{9.55J}{1.70kg}} \\ =\sqrt[]{5.6176\ldots\frac{m^2}{s^2}} \\ =2.37\frac{m}{s} \end{gathered}[/tex]Therefore, the initial speed of the block is:
[tex]2.37\frac{m}{s}[/tex]