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If a body is moved by defeating a force on a given path, energy can be stored that can then accelerate the body by imparting a speed and thereby kinetic energy. Stored energy has the potential to accelerate the body and is therefore called potential energy.

>Model

ID:(752, 0)

Equation

>Top, >Model

At the surface of the planet, the gravitational force is

and the energy

can be shown to be

$ V = m_g g z $

Conditions

Variables

$g$

Gravitational Acceleration

9.8

$m/s^2$

5310

$m_g$

Gravitational mass

$kg$

8762

$z$

Height above Floor

$m$

5286

$V$

Potential Energy

$J$

4981

Relation

Development

As the gravitational force is

$ F_g = m_g g $

with $m$ representing the mass. To move this mass from a height $h_1$ to a height $h_2$, a distance of

$ V = m g ( h_2 - h_1 )$

is covered. Therefore, the energy

$ dW = \vec{F} \cdot d\vec{s} $

with $\Delta s=\Delta h$ gives us the variation in potential energy:

$\Delta W = F\Delta s=mg\Delta h=mg(h_2-h_1)=U_2-U_1=\Delta V$

thus, the gravitational potential energy is

$ V = m_g g z $

ID:(3245, 0)

Equation

>Top, >Model

To lift an object from height $h_1$ to a height $h_2$, energy is required, which we will call gravitational potential energy

and which is proportional to the gained height:

$ V = m g ( h_2 - h_1 )$

Conditions

Variables

$g$

Gravitational Acceleration

9.8

$m/s^2$

5310

$h_1$

Height 1

$m$

7114

$h_2$

Height 2

$m$

7115

$m$

Mass

$kg$

5183

$V$

Potential Energy

$J$

4981

Relation

Development

When an object moves from a height $h_1$ to a height $h_2$, it covers the difference in height

$h = h_2 - h_1$

thus, the potential energy

$ V = m_g g z $

becomes equal to

$ V = m g ( h_2 - h_1 )$

ID:(7111, 0)

Equation

>Top, >Model

For a pendulum with length $L$ that is deflected at an angle $\theta$, the mass is raised

by a height equal to:

$ h = L (1-\cos \theta )$

Conditions

Variables

$h$

Height in Case Pendulum

$m$

6296

$L$

Pendulum Length

$m$

6282

$\theta$

Swing angle

$rad$

6283

Relation

ID:(4523, 0)

Equation

>Top, >Model

For the case of a mass $m$ hanging from a string of length $L$ and being deflected at an angle $\theta$ from the vertical, the mass will gain a height of

which means that the gravitational potential energy

will be

$ U = m g L (1-\cos \theta )$

Conditions

Variables

$g$

Gravitational Acceleration

9.8

$m/s^2$

5310

$m_g$

Gravitational mass

$kg$

8762

$L$

Pendulum Length

$m$

6282

$U$

Potential Energy Pendulum

$J$

6284

$\theta$

Swing angle

$rad$

6283

Relation

where $g$ is the acceleration due to gravity.

ID:(4513, 0)

Equation

>Top, >Model

The gravitational potential energy of a pendulum is

which for small angles can be approximated as:

$ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$

Conditions

Variables

$g$

Gravitational Acceleration

9.8

$m/s^2$

5310

$m_g$

Gravitational mass

$kg$

8762

$L$

Pendulum Length

$m$

6282

$V$

Potential Energy Pendulum, for small Angles

$J$

6285

$\theta$

Swing angle

$rad$

6283

Relation

Development

The gravitational potential energy of a pendulum with mass m, suspended from a string of length L and deflected by an angle \theta is given by

$ U = m g L (1-\cos \theta )$

where g is the acceleration due to gravity.

For small angles, the cosine function can be approximated using a Taylor series expansion up to the second term

$\cos\theta\sim 1-\displaystyle\frac{1}{2}\theta^2$

This approximation leads to the simplification of the potential energy to

$ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$

It's important to note that the angle must be expressed in radians.

ID:(4514, 0)

Equation

>Top, >Model

En el caso elástico (resorte) la fuerza es



la energía

se puede mostrar que en este caso es

$ V =\displaystyle\frac{1}{2} k x ^2$

Conditions

Variables

$x$

Elongation of the Spring

$m$

5313

$k$

Hooke Constant

$N/m$

5311

$V$

Potential Energy

$J$

4981

Relation

Development

En el caso elástico (resorte) la fuerza es



con k la constante del resorte y x la elongación/compresión del resorte. La variación de la energía potencial es

$ dW = \vec{F} \cdot d\vec{s} $

\\n\\nLa diferencia\\n\\n

$\Delta x = x_2 - x_1$

\\n\\ncorresponde al camino recorrido por lo que\\n\\n

$\Delta W=k,x,\Delta x=k(x_2-x_1)\displaystyle\frac{(x_1+x_2)}{2}=\displaystyle\frac{k}{2}(x_2^2-x_1^2)$

y con ello la energía potencial elástica es

$ V =\displaystyle\frac{1}{2} k x ^2$

ID:(3246, 0)

Equation

>Top, >Model

The elongation $\Delta x$ of a spring is calculated as the difference between its original position $x_1$ and its current position $x_2$, which is expressed as

$ V =\displaystyle\frac{1}{2} k ( x_2 ^2- x_1 ^2)$

Conditions

Variables

$k$

Hooke Constant

$N/m$

5311

$s_1$

Position 1

$m$

5481

$s_2$

Position 2

$m$

5482

$V$

Potential Energy

$J$

4981

Relation

It is commonly defined that if a spring is stretched, the elongation is positive, and if it is compressed, it is negative.

ID:(7112, 0)

Equation

>Top, >Model

The gravitational force in general is expressed as

while the energy

can be shown to be

$ V = - \displaystyle\frac{ G m M }{ r } $

Conditions

Variables

$r$

Distance to the center of the celestial body

$m$

8758

$V$

General gravitational potential energy

$-$

9792

$G$

Gravitational constant

6.673e-11

$m^3/kg s^2$

8759

$m_g$

Gravitational mass

$kg$

8762

$M$

Mass of the celestial body

$kg$

8756

Relation

Development

As the gravitational force is

$ F = G \displaystyle\frac{ m_g M }{ r ^2}$

To move a mass $m$ from a distance $r_1$ to a distance $r_2$ from the center of the planet, a potential energy is required

$ W =\displaystyle\int_C \vec{F} \cdot d \vec{s} $

resulting in the gravitational potential energy being

$W_2-W_1=\displaystyle\int_{r_1}^{r_2}\displaystyle\frac{GmM}{r^2}dr=\displaystyle\frac{GmM}{r_1}-\displaystyle\frac{GmM}{r_2}$

thus yielding

$ V = - \displaystyle\frac{ G m M }{ r } $

ID:(12551, 0)

0

Video

Video: Potential Energy