91Ó°ÊÓ

The weight of wood block is $2.0$kg

Inclined angle of ramp $30°$

Initial speed is $10$m/s

The coefficient of kinetic friction between two wood is$0.20$

${v_f}^2={v_i}^2+2a∆s$

$v_i=10$m/s and the$v_f=0$

m/s we can solve for acceleration.

Solving for summation of forces horizontally with respect to the wooden ramp that is inclined for $30°$

$$\begin{gathered} \underset{}{∑}F=0 \\ ma-f_k-mg\sin (30°)=0 \\ ma={\mathrm{μ}}_{k}mg\cos (30°)+mg\sin (30°) \\ a=g\left({\mathrm{μ}}_{k}mg\cos \right(30°)+\sin (30°\left)\right) \\ a=6.60m/s^2 \\ {v_f}^2={v_i}^2+2a∆s \\ ∆s=\frac{{v_f}^2+{v_i}^2}{2a} \\ ∆s=\frac{{\left(10\right)}^2+0^2}{2(6.60)} \\ ∆s=7.58m \end{gathered}$$

$∆s$is the distance traveled by the block

$$\begin{gathered} \sin (30°)=\frac{∆s}{h} \\ h=\sin 30°×∆s \\ h=3.79m \end{gathered}$$

Compute the acceleration going down the ramp

$$\begin{gathered} \underset{}{∑}F=0 \\ ma+mg\sin (30°)-f_k=0 \\ ma+mg\sin (30°)-{\mathrm{μ}}_{k}mg\cos (30°)=0 \\ a={\mathrm{μ}}_{k}g\cos (30°)-g\sin (30°) \\ a=g\left({\mathrm{μ}}_k\cos \right(30°)-\sin (30°\left)\right) \\ a=-3.20m/s^2 \end{gathered}$$

Let's calculate the final velocity when it slides back down

$$\begin{gathered} {v_f}^2={v_i}^2+2a∆s \\ {v_f}^2=0+2(-3.20)(7.58) \\ {v}_f=\sqrt{2\left(\left|-3.20\right|\right)(7.58)} \\ {v}_f=7.0m/s \end{gathered}$$