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Energy Forms and Transfers - Kinetic and Potential Energy

Grade 6IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Energy is defined as the capacity to do work and is measured in the SI unit Joules (JJ).

Kinetic Energy (EkE_k) is the energy possessed by an object due to its motion. It depends on the object's mass (mm) and the square of its velocity (vv).

Gravitational Potential Energy (EpE_p) is the energy stored in an object because of its vertical position or height (hh) above a reference point.

The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another. Therefore, in a closed system, the total energy (Etotal=Ek+EpE_{total} = E_k + E_p) remains constant.

Energy Transfer occurs when energy moves from one object to another (e.g., a bat hitting a ball), while Energy Transformation occurs when energy changes from one type to another (e.g., EpE_p turning into EkE_k as an object falls).

Mechanical Energy is the sum of the potential and kinetic energy in a system: Emech=Ek+EpE_{mech} = E_k + E_p.

📐Formulae

Ek=12mv2E_k = \frac{1}{2}mv^2

Ep=mghE_p = mgh

Etotal=Ek+EpE_{total} = E_k + E_p

g9.8 m/s2g \approx 9.8 \text{ m/s}^2

💡Examples

Problem 1:

A soccer ball with a mass of 0.45 kg0.45 \text{ kg} is kicked and travels at a velocity of 10 m/s10 \text{ m/s}. Calculate its kinetic energy (EkE_k).

Solution:

Ek=12×0.45 kg×(10 m/s)2=22.5 JE_k = \frac{1}{2} \times 0.45 \text{ kg} \times (10 \text{ m/s})^2 = 22.5 \text{ J}

Explanation:

We use the kinetic energy formula by substituting the mass (m=0.45m = 0.45) and velocity (v=10v = 10). Squaring the velocity gives 100100, which is then multiplied by half the mass.

Problem 2:

A 2 kg2 \text{ kg} textbook is lifted from the floor to a shelf that is 1.5 meters1.5 \text{ meters} high. Calculate the Gravitational Potential Energy (EpE_p) gained by the book. (Use g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

Ep=2 kg×9.8 m/s2×1.5 m=29.4 JE_p = 2 \text{ kg} \times 9.8 \text{ m/s}^2 \times 1.5 \text{ m} = 29.4 \text{ J}

Explanation:

The potential energy is found by multiplying the mass of the book, the acceleration due to gravity, and the height of the shelf.

Problem 3:

A roller coaster car has 5000 J5000 \text{ J} of potential energy at the top of a hill and 0 J0 \text{ J} of kinetic energy. If it slides down to a point where its potential energy is 2000 J2000 \text{ J}, how much kinetic energy does it have? (Assume no friction)

Solution:

Etotal=5000 J+0 J=5000 JE_{total} = 5000 \text{ J} + 0 \text{ J} = 5000 \text{ J}. At the new point: 5000 J=2000 J+Ek5000 \text{ J} = 2000 \text{ J} + E_k. Therefore, Ek=3000 JE_k = 3000 \text{ J}.

Explanation:

According to the Law of Conservation of Energy, the total mechanical energy must remain constant (5000 J5000 \text{ J}). As potential energy decreases, it is transformed into kinetic energy.

Kinetic and Potential Energy - Revision Notes & Key Formulas | IB Grade 6 Science