A box of mass$\mathrm{m} = 10 \mathrm{kg}$ is projected up an inclined plane from its foot with a speed of $20 \mathrm{m}/\mathrm{s}$ as shown in the figure. The coefficient of friction $\mathrm{\mu }$ between the box and the plane is $0.5$. Find the distance travelled by the box on the plane before it stops for the first time.

The box starts from position$1$ with speed ${\mathrm{v}}_1 = 20 \mathrm{m}/\mathrm{s}$ and stops at position $2$. 

Kinetic energy at position $1$:${\mathrm{K}}_1=\frac{1}{2}{\mathrm{mv}}_1^2=2000\mathrm{J}$

Kinetic energy at position $2: {\mathrm{K}}_2 = 0$

${\mathrm{W}}_{1\to 2}={\mathrm{W}}_{\mathrm{g},1\to 2}+{\mathrm{W}}_{\mathrm{f},1\to 2}=-60\mathrm{x}-40\mathrm{x}=-100\mathrm{x} \mathrm{J}$

Applying work energy theorem for the motion of the box from position $1$ to position $2$, we have

$$\begin{gathered} {\mathrm{W}}_{1\to 2}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ \Rightarrow -100\mathrm{x}=0-2000 \\ \Rightarrow \mathrm{x}=20 \mathrm{m} \end{gathered}$$