Kinetic and potential energy

Energy, the kinetic energy of a moving body, the potential energy of a raised body and of a deformed spring.

Terms

Studied 0 of 5

Formulas

Kinetic energy

E_к = \dfrac{m v^2}{2}

Quantities
Symbol	Name	Unit
$E_к$	kinetic energy	J
$m$	mass of the body	kg
$v$	speed of the body	m/s

Solved for another quantity:

m = \dfrac{2 E_к}{v^2}

v = \sqrt{\dfrac{2 E_к}{m}}

Note. For example, a body of mass $2\ \text{kg}$ moves at $3\ \text{m/s}$ : $E_k = 2\cdot 3^2 / 2 = 9\ \text{J}$ .

Potential energy of a raised body

E_п = m g h

Quantities
Symbol	Name	Unit
$E_п$	potential energy	J
$m$	mass of the body	kg
$g$	acceleration due to gravity	m/s²
$h$	height above the zero level	m

Solved for another quantity:

m = \dfrac{E_п}{g h}

h = \dfrac{E_п}{m g}

Note. $g \approx 9.8\ \text{m/s}^2$ , in problems $10\ \text{m/s}^2$ is often used. For example, a $3\ \text{kg}$ body at $5\ \text{m}$ with $g = 10$ : $E_p = 3\cdot 10\cdot 5 = 150\ \text{J}$ .

Potential energy of a deformed spring

E_п = \dfrac{k x^2}{2}

Quantities
Symbol	Name	Unit
$E_п$	potential energy of the spring	J
$k$	spring stiffness	N/m
$x$	deformation (extension or compression) of the spring	m

Solved for another quantity:

k = \dfrac{2 E_п}{x^2}

x = \sqrt{\dfrac{2 E_п}{k}}

Note. For example, a spring of stiffness $200\ \text{N/m}$ is stretched by $0.1\ \text{m}$ : $E_p = 200\cdot 0.1^2 / 2 = 1\ \text{J}$ .

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

Energies of a body and a spring

The kinetic energy and the potential energy of the raised body use the same mass m; the spring is independent of the body. Potential energy is counted from the zero level h = 0.

Graphs

E_k(v)

A parabola: twice the speed gives 4 times the energy

E_p(h)

The slope of the line is m·g

E_k

25 J

E_p\ (mgh)

60 J

E_p\ (kx^2/2)

4 J

Formulas with your numbers

E_{k} = \dfrac{m v^2}{2} = \dfrac{2\cdot 5^2}{2} = 25\ \text{J}

E_{p} = m g h = 2\cdot 10\cdot 3 = 60\ \text{J}

E_{p} = \dfrac{k x^2}{2} = \dfrac{200\cdot 0.2^2}{2} = 4\ \text{J}

How to read the graphs

The $E_k(v)$ graph is a parabola with its vertex at the origin: doubling the speed makes the kinetic energy $4$ times greater.
The $E_p(h)$ graph is a straight line through the origin: potential energy is directly proportional to the height. The slope of the line is $m g$ .
The greater the mass, the higher the parabola $E_k(v)$ and the steeper the line $E_p(h)$ .
Potential energy depends on the choice of the zero level: at the zero level $h = 0$ and $E_p = 0$ .

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Terms

Studied 0 of 5

Formulas

Kinetic energy

E_к = \dfrac{m v^2}{2}

Quantities
Symbol	Name	Unit
$E_к$	kinetic energy	J
$m$	mass of the body	kg
$v$	speed of the body	m/s

Solved for another quantity:

m = \dfrac{2 E_к}{v^2}

v = \sqrt{\dfrac{2 E_к}{m}}

Note. For example, a body of mass $2\ \text{kg}$ moves at $3\ \text{m/s}$ : $E_k = 2\cdot 3^2 / 2 = 9\ \text{J}$ .

Potential energy of a raised body

E_п = m g h

Quantities
Symbol	Name	Unit
$E_п$	potential energy	J
$m$	mass of the body	kg
$g$	acceleration due to gravity	m/s²
$h$	height above the zero level	m

Solved for another quantity:

m = \dfrac{E_п}{g h}

h = \dfrac{E_п}{m g}

Note. $g \approx 9.8\ \text{m/s}^2$ , in problems $10\ \text{m/s}^2$ is often used. For example, a $3\ \text{kg}$ body at $5\ \text{m}$ with $g = 10$ : $E_p = 3\cdot 10\cdot 5 = 150\ \text{J}$ .

Potential energy of a deformed spring

E_п = \dfrac{k x^2}{2}

Quantities
Symbol	Name	Unit
$E_п$	potential energy of the spring	J
$k$	spring stiffness	N/m
$x$	deformation (extension or compression) of the spring	m

Solved for another quantity:

k = \dfrac{2 E_п}{x^2}

x = \sqrt{\dfrac{2 E_п}{k}}

Note. For example, a spring of stiffness $200\ \text{N/m}$ is stretched by $0.1\ \text{m}$ : $E_p = 200\cdot 0.1^2 / 2 = 1\ \text{J}$ .

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

Energies of a body and a spring

The kinetic energy and the potential energy of the raised body use the same mass m; the spring is independent of the body. Potential energy is counted from the zero level h = 0.

Graphs

E_k(v)

A parabola: twice the speed gives 4 times the energy

E_p(h)

The slope of the line is m·g

E_k

25 J

E_p\ (mgh)

60 J

E_p\ (kx^2/2)

4 J

Formulas with your numbers

E_{k} = \dfrac{m v^2}{2} = \dfrac{2\cdot 5^2}{2} = 25\ \text{J}

E_{p} = m g h = 2\cdot 10\cdot 3 = 60\ \text{J}

E_{p} = \dfrac{k x^2}{2} = \dfrac{200\cdot 0.2^2}{2} = 4\ \text{J}

How to read the graphs

The

E_k(v)

graph is a parabola with its vertex at the origin: doubling the speed makes the kinetic energy

4

times greater.

The

E_p(h)

graph is a straight line through the origin: potential energy is directly proportional to the height. The slope of the line is

m g

The greater the mass, the higher the parabola

E_k(v)

and the steeper the line

E_p(h)

Potential energy depends on the choice of the zero level: at the zero level

h = 0

and

E_p = 0