The law of conservation of energy | POINT classical mechanics

* In previous Work Kinetic energy Potential Energy

we learned that a body's energy is the body's ability to do work. There are several forms of energy that can be transformed into one another, such as:

(1) Potential energy to be transformed into kinetic energy in a waterfall

(2) Chemical potential energy stored in fuel (such as Gasoline, etc.) to be transformed into mechanical work represented by the movement of cars and trains

(3) Electrical energy in a light bulb to be transformed into heat and light energy

(4) Chemical potential energy stored in a battery to be transformed into electrical energy when connected in a closed electrical circuit

(5) Chemical potential energy stored in wood to be transformed into light and heat energy when it is ignited

*When energy is transformed from one form to another, the amount of energy remains constant. This is known as the law of conservation of energy.

The law of conservation of energy: Energy cannot be created or enriched, but it can be transformed from one form to another.

*Below, we will study one form of the law of conservation of energy. It is the law of conservation of mechanical energy.

Deduction of the Law of Conservation of Mechanical Energy

*Suppose a body of mass m is thrown vertically upward from point (1) with an initial velocity of vi against the direction of gravity to reach point (2) at a speed of vf. The work done on the body by the force of gravity during its rise acts to:

1) Increase the potential energy of the body with increasing height.

2) Decrease the kinetic energy of the body with a decrease in its velocity.

From the third equation of motion: $$ v^{2}_{f}-v^{2}_{i}=2ad$$

Since the body moves upward against the direction of the gravitational field.

$$\therefore  a=-g$$

$$\therefore v^{2}_{f}-v^{2}_{i}=2ad$$

$$\textrm{By multiplying the previous equation by }(\frac{1}{2}m): $$

$$\frac{1}{2}m(v^{2}_{f}-v^{2}_{i})=-mgd$$

$$\because d=y_{f}-y_{i}$$

$$\therefore \frac{1}{2}m(v^{2}_{f}-v^{2}_{i})=mg(y_{f}-y_{i})$$

$$\frac{1}{2}mv^{2}_{f}-\frac{1}{2}mv^{2}_{i}=-mgy_{f}+mgy_{i}$$

$$ mgy_{f}+\frac{1}{2}mv^{2}_{f}=mgy_{i}+\frac{1}{2}mv^{2}_{i}$$

$${\color{Red} (PE)_{f}+(KE)_{f}=(PE)_{i}+(KE)_{i} }$$

That is mean: Total potential and kinetic energies at point (1) = Total potential and kinetic energies at point (2)

Deduction: Neglecting frictional forces, the total potential and kinetic energies of the body at any point in its path = constant value called the mechanical energy (E).

That is, whenever the kinetic energy of the body increases, this is at the expense of the potential energy (the potential energy decreases) and vice versa

*the above, mechanical energy and the law of conservation of mechanical energy can be defined as follows:

Mechanical energy: The sum of the potential and kinetic energies of a body.

and

The law of conservation of mechanical energy: The sum of the potential and kinetic energies of a body at any point in its path when stationary is called the mechanical energy. It moves under the influence of gravity and neglecting air resistance, and is equal to the amount of

Notes

(1) When a body moves vertically: Under the influence of gravity and neglecting air resistance, then:

(1.1) At the maximum height of the body:

$$\because v=0$$

$$\therefore KE=0$$

$$\therefore E=PE $$

(1.2) At the halfway point between the Earth's surface and the maximum height:

$$KE=PE$$

$$\therefore E=2KE=2PE$$

(1.3) The instant the body reaches the Earth's surface:

$$\because h=0$$

$$\therefore PE=0$$

$$\therefore E=KE$$

Thus:

$$E=PE_{(\textrm{At maximum height})}=KE_{(\textrm{At ground level})}$$

(2) In the case of a simple pendulum as shown in the figure:

At (A) the ball has made its maximum displacement from its equilibrium position, and: 

$$v=0$$

$$\therefore KE=0$$

$$\therefore E=PE$$

Point (B) represents the equilibrium position of the ball; at the point, the pendulum’s speed is at its maximum.

The height (h) represents the vertical distance between the equilibrium position and the position of the pendulum’s ball at any point in the relationship: 

$$(PE=mgh)$$