User:
Potential gravitational energy
Storyboard 
When a body is lifted against the gravitational force to a given height, it gains gravitational potential energy, which is proportional to its mass, the gravitational acceleration, and the height reached.
ID:(1422, 0)
Roller coaster
Description
If we consider a roller coaster car moving without friction, its energy will remain conserved.
This means that at any pair of points we consider, the total energy will always be the same:
| $ E_1 = E_2 $ |
Since the energy is composed of one part total Kinetic Energy ($K$) and another the potential Energy ($V$), according to:
| $ E = K + V $ |
it follows that if one component increases, the other must decrease, and vice versa. As translational Kinetic Energy ($K_t$) depends on the inertial Mass ($m_i$) and the speed ($v$), as described in:
| $ K_t =\displaystyle\frac{1}{2} m_i v ^2$ |
while the potential Energy ($V$) depends on the gravitational mass ($m_g$), the gravitational Acceleration ($g$), and the height above Floor ($z$), according to:
| $ V = m_g g z $ |
we see that whenever the height increases, the velocity will decrease, and vice versa. In this sense, we can predict the speed at any height along the roller coaster.
If you are studying a specific case, be cautious when assuming data to ensure a solution exists. If you assume an energy that is too low compared to a higher potential energy, there will be no velocity for which the equations have a solution. This would mean you are considering a position the object could never reach due to insufficient energy.
ID:(15866, 0)
Model
Top
Parameters
Variables
Calculations
to 
, then, select the variable: 
to 
Calculations
Calculations
Variable 
Given Calculate 
Target : Equation To be used 
Equations
$ E_1 = K_1 + V_1 $
E = K + V
$ E_2 = K_2 + V_2 $
E = K + V
$ E_1 = E_2 $
E_1 = E_2
$ K_1 =\displaystyle\frac{1}{2} m_i v_1 ^2$
K_t = m_i * v ^2/2
$ K_2 =\displaystyle\frac{1}{2} m_i v_2 ^2$
K_t = m_i * v ^2/2
$ m_g = m_i $
m_g = m_i
$ V_1 = m_g g z_1 $
V = m_g * g * z
$ V_2 = m_g g z_2 $
V = m_g * g * z
ID:(15863, 0)
Time invariant
Equation
Invariance (=no change) with respect to time means that something does not change as time passes. In other words, if something happens in a certain way today, it will happen the same way tomorrow.
Time invariance is associated with the conservation of energy. This implies that the sum of all energies will be equal to the total energy present at the beginning:
| $ E_1 = E_2 $ |
An example is an object in a gravitational field that consistently behaves the same way, indicating that the gravitational field does not dissipate energy from moving objects within it.
ID:(1177, 0)
Total Energy (1)
Equation
The total energy corresponds to the sum of the total kinetic energy and the potential energy:
| $ E_1 = K_1 + V_1 $ |
| $ E = K + V $ |
ID:(3687, 1)
Total Energy (2)
Equation
The total energy corresponds to the sum of the total kinetic energy and the potential energy:
| $ E_2 = K_2 + V_2 $ |
| $ E = K + V $ |
ID:(3687, 2)
Translational Kinetic Energy (1)
Equation
In the case of studying translational motion, the definition of energy
| $ dW = \vec{F} \cdot d\vec{s} $ |
is applied to Newton's second law
| $ F = m_i a $ |
resulting in the expression
| $ K_1 =\displaystyle\frac{1}{2} m_i v_1 ^2$ |
| $ K_t =\displaystyle\frac{1}{2} m_i v ^2$ |
The energy required for an object to transition from velocity $v_1$ to velocity $v_2$ can be calculated using the definition with
| $ dW = \vec{F} \cdot d\vec{s} $ |
Using the second law of Newton, this expression can be rewritten as
$\Delta W = m a \Delta s = m\displaystyle\frac{\Delta v}{\Delta t}\Delta s$
Employing the definition of velocity with
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
we obtain
$\Delta W = m\displaystyle\frac{\Delta v}{\Delta t}\Delta s = m v \Delta v$
where the difference in velocities is
$\Delta v = v_2 - v_1$
Furthermore, the velocity itself can be approximated by the average velocity
$v = \displaystyle\frac{v_1 + v_2}{2}$
Using both expressions, we arrive at
$\Delta W = m v \Delta v = m(v_2 - v_1)\displaystyle\frac{(v_1 + v_2)}{2} = \displaystyle\frac{m}{2}(v_2^2 - v_1^2)$
Thus, the change in energy is given by
$\Delta W = \displaystyle\frac{m}{2}v_2^2 - \displaystyle\frac{m}{2}v_1^2$
In this way, we can define kinetic energy
| $ K_t =\displaystyle\frac{1}{2} m_i v ^2$ |
ID:(3244, 1)
Translational Kinetic Energy (2)
Equation
In the case of studying translational motion, the definition of energy
| $ dW = \vec{F} \cdot d\vec{s} $ |
is applied to Newton's second law
| $ F = m_i a $ |
resulting in the expression
| $ K_2 =\displaystyle\frac{1}{2} m_i v_2 ^2$ |
| $ K_t =\displaystyle\frac{1}{2} m_i v ^2$ |
The energy required for an object to transition from velocity $v_1$ to velocity $v_2$ can be calculated using the definition with
| $ dW = \vec{F} \cdot d\vec{s} $ |
Using the second law of Newton, this expression can be rewritten as
$\Delta W = m a \Delta s = m\displaystyle\frac{\Delta v}{\Delta t}\Delta s$
Employing the definition of velocity with
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
we obtain
$\Delta W = m\displaystyle\frac{\Delta v}{\Delta t}\Delta s = m v \Delta v$
where the difference in velocities is
$\Delta v = v_2 - v_1$
Furthermore, the velocity itself can be approximated by the average velocity
$v = \displaystyle\frac{v_1 + v_2}{2}$
Using both expressions, we arrive at
$\Delta W = m v \Delta v = m(v_2 - v_1)\displaystyle\frac{(v_1 + v_2)}{2} = \displaystyle\frac{m}{2}(v_2^2 - v_1^2)$
Thus, the change in energy is given by
$\Delta W = \displaystyle\frac{m}{2}v_2^2 - \displaystyle\frac{m}{2}v_1^2$
In this way, we can define kinetic energy
| $ K_t =\displaystyle\frac{1}{2} m_i v ^2$ |
ID:(3244, 2)
Gravitational potential energy at the planet's surface (1)
Equation
At the surface of the planet, the gravitational force is
| $ F_g = m_g g $ |
and the energy
| $ dW = \vec{F} \cdot d\vec{s} $ |
can be shown to be
| $ V_1 = m_g g z_1 $ |
| $ V = m_g g z $ |
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, 1)
Gravitational potential energy at the planet's surface (2)
Equation
At the surface of the planet, the gravitational force is
| $ F_g = m_g g $ |
and the energy
| $ dW = \vec{F} \cdot d\vec{s} $ |
can be shown to be
| $ V_2 = m_g g z_2 $ |
| $ V = m_g g z $ |
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, 2)
Equality of inertial and gravitational mass
Equation
The masses that Newton used in his principles are related to the inertia of bodies, which leads to the concept of the inertial Mass ($m_i$).
Newton's law, which is linked to the force between bodies due to their masses, is related to gravity, hence known as the gravitational mass ($m_g$).
Empirically, it has been concluded that both masses are equivalent, and therefore, we define
| $ m_g = m_i $ |
Einstein was the one who questioned this equality and, from that doubt, understood why both 'appear' equal in his theory of gravity. In his argument, Einstein explained that masses deform space, and this deformation of space causes a change in the behavior of bodies. Thus, masses turn out to be equivalent. The revolutionary concept of space curvature implies that even light, which lacks mass, is affected by celestial bodies, contradicting Newton's theory of gravitation. This was experimentally demonstrated by studying the behavior of light during a solar eclipse. In this situation, light beams are deflected due to the presence of the sun, allowing stars behind it to be observed.
ID:(12552, 0)