Answer
$ v_f =2.7m/s$
Work Step by Step
$d = 0.55m + 0.13$
$\Delta K + E_{th} = U$
$ (K_f -K_i) + E_{th} = U$
$ (\frac{1}{2}mv^2_f -\frac{1}{2}mv^2_i ) + fd = mgdsin\theta$
$ \frac{1}{2}m(v^2_f -v^2_i ) + fd = mgdsin\theta$
The initial velocity$(v_i)=0$ because the jar is moved from rest
$ \frac{1}{2}m(v^2_f -0 ) + fd = mgdsin\theta$
$ \frac{1}{2}mv^2_f + fd = mgdsin\theta$
$ \frac{1}{2}mv^2_f + (μ_k F_N)d = mgdsin\theta$
$ \frac{1}{2}mv^2_f + (μ_k mgcos\theta )d = mgdsin\theta$
$ \frac{1}{2}mv^2_f = mgdsin\theta - μ_k mgdcos\theta $
$ v^2_f = 2\times \frac{mgdsin\theta - μ_k mgdcos\theta}{m} $
$ v^2_f = 2gd(sin\theta - μ_k cos\theta) $
$ v_f =\sqrt { 2gd(sin\theta - μ_k cos\theta)} $
$ v_f =\sqrt { (2)(9.8)(0.68)(sin40^{\circ}- (0.15 \times cos40^{\circ}))} $
$ v_f =2.7m/s$
