A cookie jar is moving up an $\mathbf{40}\mathbf{°}$incline. At a point 55 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.4 m/s . The coefficient of kinetic friction between jar and incline is 0.15 . (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?

The distance travelled by the jar up the incline${\mathrm{d}}_1=55\mathrm{cm}=0.55\mathrm{m}$

The speed of the block at distance${\mathrm{d}}_1$up the incline,${\mathrm{v}}_0=1.4\mathrm{m}/\mathrm{s}$

The angle of the incline with the horizontal = 40⁰

The coefficient of friction,${\mathrm{\mu }}_{\mathrm{k}}=0.15$

The jar moving up the incline gains potential energy. At a given point on the incline, it has acquired kinetic energy as well as potential energy. But it loses some of its energy to the friction between the jar and the surface. Hence, after reaching a certain distance, it will stop. When it starts sliding down again, its potential energy will be converted partly to kinetic energy and some energy will be spent as thermal energy. Thus, we use the principle of conservation of energy to determine the distance travelled and the speed of the jar.

Formula:

$\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$

$$\begin{gathered} \mathrm{P}.\mathrm{E}=\mathrm{mgh} \\ {\mathrm{E}}_{\mathrm{th}}={\mathrm{F}}_{\mathrm{f}}\mathrm{d} \end{gathered}$$

Using the diagram of the incline, we can write,

$$\begin{gathered} \sin 40=\frac{\mathrm{h}}{{\mathrm{d}}_2} \\ \mathrm{h}={\mathrm{d}}_2\sin 40 \end{gathered}$$

And

N = mg cos40

Now, using the conservation of energy principle,

$\left(\mathrm{K}.\mathrm{E}.\right)\mathrm{at}\mathrm{bottom}=\left(\mathrm{P}.\mathrm{E}\right)\mathrm{at}\mathrm{h}+{\mathrm{E}}_{\mathrm{th}}$

localid="1661399464250" $$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}+\mathrm{\mu }\mathrm{N}{\mathrm{d}}_2 \\ \frac{1}{2}{\mathrm{mv}}^2=\mathrm{mg}\sin 40+\mathrm{\mu }\mathrm{mg}\cos 40{\mathrm{d}}_2 \end{gathered}$$

Simplify by cancelling out m and collecting similar terms together to get

$$\begin{gathered} \left(\sin 40+\mathrm{\mu }\cos 40\right){\mathrm{d}}_2=\frac{{\mathrm{v}}^2}{2\mathrm{g}} \\ \left(0.64+\left(0.15\times 0.77\right)\right){\mathrm{d}}_2=\frac{1.96}{2\times 9.8} \\ {\mathrm{d}}_2=\frac{0.1}{0.76} \\ =0.13\mathrm{m} \end{gathered}$$

Hence, the distance travelled by the cookie is 0.13 m

Now, the total distance travelled by the jar on the incline is

$$\begin{gathered} \mathrm{d}={\mathrm{d}}_1+{\mathrm{d}}_2 \\ =0.55+0.13 \\ =0.68\mathrm{m} \end{gathered}$$

So, when the cookie starts sliding back to the ground, the conservation of energy law gives

$$\begin{gathered} \mathrm{P}.{\mathrm{E}}_{\mathrm{top}}={\mathrm{E}}_{\mathrm{th}}+\mathrm{K}.{\mathrm{E}}_{\mathrm{f}} \\ \mathrm{K}.{\mathrm{E}}_{\mathrm{f}}=\mathrm{P}.{\mathrm{E}}_{\mathrm{top}}-{\mathrm{E}}_{\mathrm{th}} \end{gathered}$$

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2=\mathrm{mgh}-\mathrm{\mu }\mathrm{N}\mathrm{d} \\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2=\mathrm{mg}\mathrm{d}\sin -40\mathrm{\mu }\mathrm{mg}\cos 40\mathrm{d} \\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2=\mathrm{gd}\left(\sin 40-\mathrm{\mu }\cos 40\right) \\ {\mathrm{v}}_{\mathrm{f}}^2=2\times 9.8\times 0.68\left[0.64-\left(0.15\times 0.77\right)\right] \\ {\mathrm{v}}_{\mathrm{f}}^2=2.65\mathrm{m}/\mathrm{s} \\ =2.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the final speed of the cookie is 2.7 m/s

When the coefficient of friction decreases, the energy loss as a result of friction will be less. Hence, the jar will travel more distance on the incline before coming to rest.

As the jar can reach a higher distance, its potential energy is also more. This results in more kinetic energy of the jar when it reaches the bottom. This, in turn, gives more speed to the jar.