Answer
$ d_f = 0.13m $
Work Step by Step
$\Delta K + \Delta E_{th} = -\Delta U$
$ (K_f -K_i) + \Delta E_{th} = - \Delta U$
$ (\frac{1}{2}mv^2_f -\frac{1}{2}mv^2_i ) + (fd_f -fd_i) = - (mgd_fsin\theta -mgd_isin\theta)$
$ \frac{1}{2}m(v^2_f -v^2_i ) + f(d_f -d_i) = - mgsin\theta(d_f-d_i)$
The comes to a stop at an unknown distance$(d_f)$ so $v_f = 0$
Since we are calculating the distance the jar will move further up the incline
and not the whole length of the incline, the initial distance $(d_i)$ will be equal to zero
$ \frac{1}{2}m(0 -v^2_i ) + f(d_f -0) = - mgsin\theta(d_f-0)$
$ -\frac{1}{2}mv^2_i + fd_f = - mgd_fsin\theta$
$ -\frac{1}{2}mv^2_i + (μ_k F_N)d_f = - mgd_fsin\theta$
$ -\frac{1}{2}mv^2_i + (μ_k mgcos\theta )d_f = - mgd_fsin\theta$
$ (μ_k mgcos\theta )d_f +mgd_fsin\theta = \frac{1}{2}mv^2_i $
$ mgd_f (μ_k cos\theta +sin\theta )= \frac{1}{2}mv^2_i $
$ \frac{mgd_f (μ_k cos\theta +sin\theta )}{m}= \frac{1}{2}v^2_i $
$ gd_f (μ_k cos\theta +sin\theta ))= \frac{1}{2} v^2_i $
$ d_f = \frac{v^2_i}{2g(μ_k cos\theta +sin\theta )} $
$ d_f = \frac{1.4^2_i}{(2)(9.8)(0.15\times cos40^{\circ} +sin40^{\circ} )} $
$ d_f = 0.13m $
