Inclined Plane Example

A mountain climber who is climbing a mountain in the Drakensberg during winter, by mistake drops her water bottle which then slides \(\text{100}\) \(\text{m}\) down the side of a steep icy slope to a point which is 10 m lower than the climber's position. The mass of the climber is \(\text{60}\) \(\text{kg}\) and her water bottle has a mass of \(\text{500}\) \(\text{g}\) .

1. If the bottle starts from rest, how fast is it travelling by the time it reaches the bottom of the slope? (Neglect friction.)

2. What is the total change in the climber's potential energy as she climbs down the mountain to fetch her fallen water bottle? i.e. what is the difference between her potential energy at the top of the slope and the bottom of the slope?

Step 1: Analyse the question to determine what information is provided

• the distance travelled by the water bottle down the slope, \(d = \text{100}\text{ m}\)

• the difference in height between the starting position and the final position of the water bottle is \(h = \text{10}\text{ m}\)

• the bottle starts sliding from rest, so its initial velocity is \({v}_{1} = \text{9.8}\text{ m·s$^{-1}$}\)

• the mass of the climber is \(\text{60}\) \(\text{kg}\)

• the mass of the water bottle is \(\text{500}\) \(\text{g}\). We need to convert this mass into \(\text{kg}\): \(\text{500}\text{ g} = \text{0.5}\text{ kg}\)

Step 2: Analyse the question to determine what is being asked

• What is the velocity of the water bottle at the bottom of the slope?

• What is the difference between the climber's potential energy when she is at the top of the slope compared to when she reaches the bottom?

Step 3: Calculate the velocity of the water bottle when it reaches the bottom of the slope

\begin{align*} {E}_{M1} & = {E}_{M2} \\ {E}_{K1} + {E}_{P1} & = {E}_{K2} + {E}_{P2} \\ \frac{1}{2}m{\left({v}_{1}\right)}^{2} + mg{h}_{1} & = \frac{1}{2}m{\left({v}_{2}\right)}^{2} + mg{h}_{2} \\ 0 + mg{h}_{1} & = \frac{1}{2}m{\left({v}_{2}\right)}^{2} + 0 \\ {\left({v}_{2}\right)}^{2} & = \frac{2mgh}{m} \\ {\left({v}_{2}\right)}^{2} & = 2gh \\ {\left({v}_{2}\right)}^{2} & = \left(2\right)\left(\text{9.8}\text{ m·s$^{-2}$}\right)\left(\text{10}\text{ m}\right) \\ {v}_{2} & = \text{14}\text{ m·s$^{-1}$} \end{align*}

Note: the distance that the bottle travelled (i.e. \(\text{100}\) \(\text{m}\)) does not play any role in calculating the energies. It is only the height difference that is important in calculating potential energy.

Step 4: Calculate the difference between the climber's potential energy at the top of the slope and her potential energy at the bottom of the slope

At the top of the slope, her potential energy is:

\begin{align*} {E}_{P1} & = mg{h}_{1} \\ & = \left(\text{60}\text{ kg}\right)\left(\text{9.8}\text{ m·s$^{-2}$}\right)\left(\text{10}\text{ m}\right) \\ & = \text{5 800}\text{ J} \end{align*}

At the bottom of the slope, her potential energy is:

\begin{align*} {E}_{P1} & = mg{h}_{1} \\ & = \left(\text{60}\text{ kg}\right)\left(\text{9.8}\text{ m·s$^{-2}$}\right)\left(\text{0}\text{ m}\right) \\ & = \text{0}\text{ J} \end{align*}

Therefore the difference in her potential energy when moving from the top of the slope to the bottom is:

\[{E}_{P1} - {E}_{P2} = \text{5 880} - 0 = \text{5 880}\text{ J}\]