Answer:
[tex]v=15.9499m/s[/tex]
Explanation:
From the question we are told that:
Mass of wood [tex]m=1kg[/tex]
force constants [tex]k= 200N-m[/tex]
Coefficient of kinetic friction [tex]\mu= 0.2[/tex]
Bullet mass [tex]m_b= 20 \approx 0.02kg[/tex]
Spring compresion [tex]y=15cm \approx 0.15 m[/tex]
Generally the equation for kinetic energy of bullet [tex]K>E_b[/tex] is mathematically given by
Complete Question
[tex]K.E_b=spring potential energy+work done against friction[/tex]
[tex]K.E_b=\frac{1}{2} mbv^2[/tex]
[tex]\frac{1}{2} m_b v^2=\frac{1}{2} ky^2+\mu my[/tex]
[tex]\frac{1}{2} (0.02)v^2=\frac{1}{2} (0.2)(0.15)^2+0.2(1)(0.15)[/tex]
[tex]v=15.9499m/s[/tex]
[tex]v\approx16m/s[/tex]