User:
Inclined plane
Storyboard 
When a body is placed on an inclined plane, it begins to slide under the action of gravity. However, its vertical velocity component is smaller than in free fall because part of the acceleration is projected onto the direction parallel to the plane, reducing its velocity in the vertical axis.
ID:(752, 0)
Potential Energy
Description
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.
Variables
Calculations
to 
,
then, select the variable: 
to 
Calculations
Variable
Given
Calculate
Target :
Equation
To be used
Equations
The energy required for an object to change its angular velocity from $\omega_1$ to $\omega_2$ can be calculated using the definition
| $ \Delta W = T \Delta\theta $ |
Applying Newton's second law, this expression can be rewritten as
$\Delta W=I \alpha \Delta\theta=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta$
Using the definition of angular velocity
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
we get
$\Delta W=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta=I \omega \Delta\omega$
The difference in angular velocities is
$\Delta\omega=\omega_2-\omega_1$
On the other hand, angular velocity itself can be approximated with the average angular velocity
$\omega=\displaystyle\frac{\omega_1+\omega_2}{2}$
Using both expressions, we obtain the equation
$\Delta W=I \omega \Delta \omega=I(\omega_2-\omega_1)\displaystyle\frac{(\omega_1+\omega_2)}{2}=\displaystyle\frac{I}{2}(\omega_2^2-\omega_1^2)$
Thus, the change in energy is given by
$\Delta W=\displaystyle\frac{I}{2}\omega_2^2-\displaystyle\frac{I}{2}\omega_1^2$
This allows us to define kinetic energy as
| $ K_t =\displaystyle\frac{1}{2} m_i v ^2$ |
(ID 3244)
(ID 3687)
(ID 12552)
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 g s \sin \phi $ |
(ID 12925)
Examples
(ID 16248)
When a body is placed on an inclined plane and there is no friction preventing it from sliding, it begins to accelerate under the action of gravity. However, the gravitational force acting vertically decomposes into a component parallel to the plane, whose magnitude is:
$F_p = m_g g \sin\theta$
This depends on the gravitational mass ($m_g$), the gravitational Acceleration ($g$), and the angle of the inclined plane ($\phi$). This force gives rise to the potential energy:
| $ V = m_g g s \sin \phi $ |
expressed as a function of the path traveled on the inclined plane ($s$).
(ID 16247)
(ID 16249)
ID:(752, 0)