Answer:
6.21 m/s
Explanation:
Using work energy equation then
[tex]U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})[/tex]
where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity
Substituting 1 Kg for m, 0.4m for h, [tex]v_a[/tex] as 0, 9.81 for g then
[tex]58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s[/tex]
