A $2.0$kg wood block is launched up a wooden ramp that is inclined at a $30°$angle. The block’s initial speed is $10$m/s.

a. What vertical height does the block reach above its starting point?

b. What speed does it have when it slides back down to its starting point?

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 $30°$

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

$∆s$is the distance traveled by the block

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

Compute the acceleration going down the ramp

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

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

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