Kinetic energy and gravitational potential energy

In a nutshell

Kinetic energy is the energy of an object associated with its motion. An object's gravitational potential energy (gpe) is its capacity to do work due to its position in a gravitational field. When an object moves up and down in a gravitational field, kinetic and gravitational potential energy exchange.

Equations

DESCRIPTION	EQUATION
Kinetic energy equation	$E_k = \frac{1}{2} mv^2$
Gravitational potential energy equation	$E_p = mgh$

Constants

coNstant	symbol	Value
Acceleration due to gravity	$g$	$9.81 \ ms^{-2}$

Variable definitions

QUANTITY NAME	SYMBOL	DERIVED UNIT	SI UNIT
$mass$	$m$	$kg$	$kg$
$height$	$h$	$m$	$m$
$velocity$	$v$	$ms^{-1}$	$ms^{-1}$

Kinetic energy

Kinetic energy is the energy of an object associated with its motion. It can be calculated using the equation:

$E_k = \frac{1}{2} mv^2$

Example

A man is running at a speed of $5 ms^{-1}$ and weighs $65 \ kg$ . Calculate the kinetic energy of the man to 2sf.

Firstly, state the variables in the question:

$v = 5 \ ms^{-1}, \ \ m=6 5 \ kg$

Then state the equation needed:

$E_k = \frac{1}{2} mv^2$

Finally, substitute in and solve:

$E_k = \frac{1}{2} \times 65 \times (5)^2\newline \\[0.1in] E_k = 812.5 \ J \newline \\[0.1in] E_k = 810 \ J \ (2sf)$

The kinetic energy of the man is $\underline{810 \ J}$ .

Gravitational potential energy

An object's gravitational potential energy (gpe) is its capacity to do work due to its position in a gravitational field. The equation for gpe assumes the Earth's gravitational field is uniform:

$W = Fx$

where:

$F= mg, x=h\newline \\[0.1in] E_p = mgh$

Note: gravity is an attractive force, so when an object gets higher in a gravitational field it gains gpe. This equation only applies on the surface of planets, where the gravitational field can be assumed to be uniform.

Example:

A bird of mass $3 \ kg$ takes off from the ground and reaches a height of $10 \ m$ . Calculate its gain in gravitational potential energy.

Firstly, state the variables in the question:

$h = 10 \ m, \ \ m=3 \ kg$

Then state the equation needed:

$E_p = mgh$

Finally, substitute in and solve:

$E_p = 3 \times 9.81 \times 10\newline \\[0.1in] E_p = 294.3 \ J\newline \\[0.1in] E_p = 290 \ J \ (2sf)$ 

The gravitational potential energy of the bird is $\underline{290 \ J}$ .

Energy exchange

When an object moves up and down in a gravitational field, its kinetic and gravitational energy exchange. If you consider a closed loop, where all kinetic energy is transferred to potential energy, an equation for the velocity of an object can be derived. As the object rises, kinetic energy is converted to gravitational potential energy, then when it falls this is converted back to kinetic energy.

Tip: If an object is travelling with horizontal velocity, it will still have a component of kinetic energy at max height.

As $E_p = E_k$ :

$mgh = \frac{1}{2} mv^2\newline \\[0.1in] 2gh = v^2\newline \\[0.1in] v = \sqrt{2gh}$

Note: the mass of an object has no effect on the final speed, as the acceleration in free fall is the same for all objects.

Physics; Energy; KS5 Year 12; Kinetic energy and gravitational potential energy

1.	Rollercoaster at height $h$ , with $E_p$
2.	$E_p$ transfers to $E_k$ as height decreases and velocity increases
3.	$v = \sqrt{2gh}$

Example

A rollercoaster cart is on the top of a hill of height $50 \ m$ calculate the the speed of the cart at the bottom of the hill to 2sf.

Firstly, state the variables in the question:

$h = 50 \ m$

Then state the equation needed:

$v = \sqrt{2gh}$

Finally, substitute in and solve:

$v = \sqrt{2\times 9.81 \times 50}\newline \\[0.1in] v =31.32 \ ms^{-1}\newline \\[0.1in] v =31 \ ms^{-1} \ (2sf)$ 

The speed of the cart at the bottom of the hill is $\underline{31 \ ms^{-1}}$ .