Gravitational Potential Energy

If a person weighing$75\mathrm{kg}$climbs a flight of stairs to reach a height of$100\mathrm{m}$then calculate:

(i) Their increase in$E_{GPE}$.

(ii) The work done by the person to climb the flight of stairs.

The work done to climb the stairs is equal to the change in gravitational potential energy, StudySmarter Originals

First, we need to calculate the increase in gravitational potential energy when the person climbs the stairs. This can be found using the formula we discussed above.

$\begin{array}{lcl}{E}_{GPE}& =& mgh\\ & =& 75\mathrm{kg}\times 100\mathrm{m}\times 9.8\mathrm{N}/\mathrm{kg}\\ & \mathbf{=}& \mathbf{73500}\mathbf{}\mathbf{J}\mathbf{}\mathbf{or}\mathbf{}\mathbf{735}\mathbf{}\mathbf{kJ}\end{array}$

Work done to climb the stairs:

We already know that the work done is equal to the potential energy gained when the person climbs to the top of the stairs.

$\mathrm{work}=\mathrm{force}\mathrm{x}\mathrm{distance}=E_{GPE}\mathbf{}\mathbf{=}\mathbf{}\mathbf{735}\mathbf{}\mathbf{kJ}$

The person does$735\mathrm{kJ}$work of to climb to the top of the stairs.

How many stairs would a person weighing$54\mathrm{kg}$need to climb to burn$2000\mathrm{calories}$? The height of each step is$15\mathrm{cm}$.

We first need to convert the units into the ones used in the equation.

Unit conversion:

$\begin{array}{rcl}1000\mathrm{calories}& =& 4184\mathrm{J}\\ 2000\mathrm{calories}& =& 8368\mathrm{J}\\ 15\mathrm{cm}& =& 0.15\mathrm{m}\end{array}$

First, we calculate the work done when a person climbs one step.

$$\begin{gathered} mgh=54\mathrm{kg}\times 9.8\mathrm{N}/\mathrm{kg}\times 0.15\mathrm{m} \\ =\mathbf{79}\mathbf{.}\mathbf{38}\mathbf{}\mathbf{J} \end{gathered}$$

Now, we can calculate the number of steps one has to scale in order to burn$2000\mathrm{calories}$or$8368\mathrm{J}$:

$$\begin{gathered} Noofsteps=\frac{8368\mathrm{J}\times 1000}{79.38\mathrm{J}} \\ =105,416steps \end{gathered}$$

A person weighing$54\mathrm{kg}$would have to climb$105,416\mathrm{steps}$to burn$2000\mathrm{calories}$, phew!

If a$500\mathrm{g}$apple is dropped from a height of$100\mathrm{m}$above the ground, at what speed will it hit the ground? Ignore any effects from air resistance.

The speed of a falling apple increases as it is accelerated by gravity, and is at a maximum at the point of impact, StudySmarter Originals

The gravitational potential energy of the object is converted into kinetic energy as it falls and increases in velocity. Therefore the potential energy at the top is equal to the kinetic energy at the bottom at the time of impact.

The total energy of the apple at all times is given by:

${\mathrm{E}}_{\mathrm{total}}={\mathrm{E}}_{\mathrm{GPE}}+{\mathrm{E}}_{\mathrm{KE}}$

When the apple is at a height of$100\mathrm{m}$, the velocity is zero hence the$E_{KE}=0$. Then the total energy is:

When the apple is about to hit the ground the potential energy is zero, hence the total energy is now:

$E_{\mathrm{total}}=E_{\mathrm{KE}}$

Velocity during impact can be found by equating the$E_{GPE}$to$E_{KE}$. At the moment of impact, the kinetic energy of the object will be equal to the potential energy of the apple when it was dropped.

$\begin{array}{rcl}mgh& =& \frac{1}{2}m{v}^2\\ gh& =& \frac{1}{2}{v}^2\\ v& =& \sqrt{2gh}\\ v& =& \sqrt{2\times 9.8\mathrm{N}/\mathrm{kg}\times 100\mathrm{m}}\\ v& =& 44.27\mathrm{m}/\mathrm{s}\end{array}$

The apple has a velocity of$44.27\mathrm{m}/\mathrm{s}$when it hits the ground.

A small frog of mass$30\mathrm{g}$jumps over a rock of height$15\mathrm{cm}$. Calculate the change in$E_{\mathrm{PE}}$for the frog, and the vertical speed at which the frog jumps to complete the leap.

The potential energy of a frog is constantly changing during a jump. It is zero at the moment the frog jumps and increases until the frog reaches its maximum height, where the potential energy is also maximum. After this, potential energy goes on to decrease as it is converted into kinetic energy of the falling frog. StudySmarter Originals

The change in energy of the frog as it makes the leap can be found as follows:

$\begin{array}{rcl}∆\mathrm{E}& =& 0.15\mathrm{m}\mathrm{x}0.03\mathrm{kg}\mathrm{x}9.8\mathrm{N}/\mathrm{kg}\\ & =& 0.0066\mathrm{J}\end{array}$

To calculate the vertical speed at take-off, we know that the total energy of the frog at all times is given by:

$E_{\mathrm{total}}=E_{\mathrm{GPE}}+E_{\mathrm{KE}}\mathbf{}\mathbf{}$

When the frog is about to jump, it's potential energy is zero, hence the total energy is now

$E_{\mathrm{total}}=E_{\mathrm{KE}}$

When the frog is at a height of$0.15\mathrm{m}$, then the total energy is in the gravitational potential energy of the frog:

$E_{total}=E_{GPE}$

The vertical velocity at the start of the jump can be found by equating the$E_{GPE}$to$E_{KE}$.

$$\begin{gathered} mgh=1/2m{v}^2 \\ gh=1/2v^2 \\ v=\sqrt{\left(2gh\right)} \\ v=\sqrt{(2X9.8\mathrm{N}/\mathrm{kg}X0.15\mathrm{m})} \\ v=1.71\mathrm{m}/\mathrm{s} \end{gathered}$$

The frog jumps with an initial vertical velocity of$1.71\mathrm{m}/\mathrm{s}$.