Answer
$v_f= 5.46m/s$
Work Step by Step
Finding acceleration:
$F_{net}=Fgsin \theta -f $
$ma=mgsin \theta - μ_kF_N$
$ma=mgsin \theta - μ_kmgcos\theta$
$a=\frac{mgsin \theta - μ_kmgcos\theta}{m}$
$a=gsin \theta - μ_kgcos\theta$
$a=gsin \theta - μ_kgcos\theta$
$a=g(sin \theta - μ_kcos\theta)$
$a=9.8\times (sin 20^{\circ} - (0.1\times cos20^{\circ}))$
$a=2.43m/s^2$
Using acceleration to get the final velocity:
$v^2_f= v_i^2 +2ad$
$v_f= \sqrt {v_i^2 +2ad}$
$v_f= \sqrt {0.457^2 +(2\times 2.43 \times 6.1)}$
$v_f= 5.46m/s$
