A 4.0 kgbundle starts up an$\mathbf{30}\mathbf{°}$incline with 128 Jof kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is 0.30?

The mass of the bundle = 4.0 kg

The initial kinetic energy of the bundle = 128 J

The angle of the incline with the horizontal = 30⁰

The coefficient of friction${\mathrm{\mu }}_{\mathrm{k}}=0.30$

The bundle starts from the ground and moves up the incline. This results in conversion of its initial kinetic energy into potential energy but with some loss of energy. The energy lost is due friction with incline surface, and hence, it stops after travelling some distance up the incline. Thus, we can apply the law of conservation of energy to determine the distance travelled by the bundle up the incline.

Formula:

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{P}.\mathrm{E}=\mathrm{mgh} \\ {E}_{th}=F_{f}d \end{gathered}$$

Using the diagram of the incline, we can write,

$$\begin{gathered} \sin 30=\frac{h}{d} \\ h=d\sin 30 \end{gathered}$$

And

$N=mg\cos 30$

Now, using the conservation of energy principle,

$$\begin{gathered} \left(\mathrm{KE}\right)\mathrm{at}\mathrm{bottom}=\left(\mathrm{PE}\right)\mathrm{at}\mathrm{h}+{\mathrm{E}}_{\mathrm{th}} \\ =\mathrm{mgh}+\mathrm{\mu }\mathrm{N}\mathrm{d} \\ =\mathrm{mg}\sin 30+\mathrm{\mu }\mathrm{mg}\cos 30\mathrm{d} \\ 128=\mathrm{mg}\left(\sin 30+\mathrm{\mu }\cos 30\right)\mathrm{d} \\ 128=4.0\times 9.8\left[0.5+\left(0.30\times 0.866\right)\right]\mathrm{d} \\ \mathrm{d}=\frac{128}{29.78} \\ =4.3\mathrm{m} \end{gathered}$$

Hence, the distance travelled by the bundle up the incline is 4.3 m