Answer:
[tex]x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}[/tex]
Explanation:
From the law of conservation of energy
Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr
[tex]0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x[/tex]
[tex]x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}[/tex]
[tex]x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}[/tex]
The required distance from A to B is [tex]x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}[/tex]