Answer:
v=0.816 m/s
Explanation:
The force of the spring and the motion of the block are in equilibrium so without any force of friction the motion is
[tex]E_{s}=E_{k}[/tex]
[tex]\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2[/tex]
First determinate the constant of the spring that produce the kinetic energy of the bloc
[tex]k=\frac{m*v^2}{d_{s}^2}[/tex]
[tex]k=\frac{5.17kg*(1.30\frac{m}{s})^2}{0.102^2m}[/tex]
[tex]k=839.8 \frac{kg}{s^2}[/tex]
Now the motion with the force of friction in the kinetic
[tex]E_{s}=E_{k}-W_{k}[/tex]
[tex]\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2-u*m*g[/tex]
Resolve to v
[tex]v^2=\frac{(k*d_{s}^2)+(2*g*u)}{m}[/tex]
[tex]v=\sqrt{\frac{839.8*(0.102m)^2-2*9.8*0.270}{5.17}}[/tex]
[tex]v=0.81 \frac{m}{s}[/tex]
