Mechanical Energy is the energy associated with the motion or position of an object, combining kinetic and potential energy. The two types of mechanical energy are:

1. Kinetic energy
2. Potential energy

Kinetic energy is the energy possessed by an object due to its motion. An example of kinetic energy is the energy possessed by a stone shot from a catapult which is able to kill a bird. The amount of kinetic energy depends on both its mass and velocity, and it can be calculated using the formula
$$ KE = \frac{1}{2} m v^2 $$
where \( m \) is the mass
and \( v \) is the velocity of the object.

Potential energy is the stored energy an object possesses due to its position or state in a force field. A mango fruit falling from a tree has the ability to break a glass beneath it due to the potential energy stored by the fruit when at rest. It can be converted into kinetic energy when the object is set in motion.

Forms of Potential Energy:

• Gravitational Potential Energy: Associated with an object's height relative to a reference point. Example: a raised weight. $${PE_{grav}} = mgh $$
• Elastic Potential Energy: Stored in objects that can be stretched or compressed. Example: a stretched spring. $$ {PE_{elastic}} = \frac{1}{2}ke² = \frac{1}{2}Fe $$
• Chemical Potential Energy: Stored in the chemical bonds of substances. Example: a charged battery.
• Phosphates: Mined for fertilizer production.
• Heavy Metals (e.g., Copper, Iron): Extracted for manufacturing and construction.
• Rock Salt: Mined for various industrial and domestic uses.

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another. According to the principle of conservation of energy, the total energy of an isolated system remains constant. This means that energy can transform from one form to another, but the total amount remains unchanged.

Consider a fruit falling from a tree. As it falls, it experiences a transformation of energy from kinetic energy (\(KE\)) to gravitational potential energy (\(PE_{\text{grav}}\)). Consider three points A, B and C describing the motion of the falling fruit from the tree

At point A, the fruit of mass m is on the tree at a height, h above the ground. The total energy possessed by fruit is \(PE_{\text{grav}}\) since it possesses no velocity $$ v = 0 $$ $$ KE = 0 $$ $$ \text{Total energy }= KE + PE = mgh $$

At point B, the fruit is falling down with a velocity v, and having a height h, above the ground. The total energy possessed by the body is the sum of KE and PE_grav $$\text{Total energy} = KE + {PE_{grav}} $$ $$ \text{Total energy} = \frac{1}{2}mv² + mgh $$

At Point C, the fruit has gotten to the ground. The fruit still possesses velocity with which it hits the ground but has no height above the ground. At point C, velocity is maximum. Hence the total energy attained by the fruit is kinetic energy only. $$ h = 0 $$ $$ P.E = 0 $$ $$ \text{Total energy} = PE + KE $$ $$ \text{Total energy} = \frac{1}{2}mv² $$

$$\text {Therefore, from point A to point C} $$ $$ PE = KE $$ $$\text{At point C, where velocity is max} $$ $$ mgh = \frac{1}{2}mv²_{max} $$ $$ 2mgh = mv²_{max} $$ $$ 2gh = v²_{max} $$ $$ v_{max} = \sqrt{2gh} $$ $$ \text{Also, h} = \frac{V²_{max}}{2g} $$

Note: For a swinging simple pendulum, KE is max at midpoint ay zero at both ends while P.E is max at the end where it has the highest displacement, and zero at the midpoint

Example 1: An object of mass 0.25kg moves at a height h, above the ground with a speed of 4m/s. If its mechanical energy at this height is 12J, determine the value of h (g = 10m/s²) (WAEC)

Example 2: An object of mass 100g projected vertically upwards from the ground level has a velocity of 20m/s at a height of 10m. Calculate its initial kinetic energy at the ground level (g = 10m/s²) (JAMB)

Example 3: A ball of mass 200g falls from a height of 5m on to a hard floor and rebounds to a height of 3m. What energy is lost by the ball as a result of the impact on the floor (g = 10m/s²) (NECO)

Example 4: A body of mass 5kg falls from a height of 10m above the ground. What is the kinetic energy of the body just before it strikes the ground? [Neglect energy losses and take g as 10m/s²](WAEC)

Example 5: A catapult is used to project a stone of mass 50g. If the rubber of the catapult has an elastic constant of 200N/m and was stretched 5cm, what speed can it give to the mass?

From the law of conservation of energy; $$ P.E = K.E $$ The energy stored in an elastic material is the elastic potential energy $$P.E_{elastic} = \frac{1}{2}ke² $$ $$ K = 200N/m $$ $$ e = 5cm = 0.05m $$ $$ m = 50g = 0.05kg $$ $$ P.E_{elastic} = K.E $$ $$ \frac{1}{2}×200× 0.05² = \frac{1}{2}× 0.05 × v² $$ $$ 0.25 = 0.025v² $$ $$ v² = \frac{0.25}{0.025} $$ $$ v² = 10 $$ $$ v = \sqrt{10} $$ $$ v = 3.16m/s $$