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Energy - Kinetic and Potential Energy

Grade 9IB

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

🔑Concepts

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

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Kinetic Energy (EkE_k) is the energy possessed by an object due to its motion. It is directly proportional to the mass (mm) and the square of the velocity (v2v^2).

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Gravitational Potential Energy (EpE_p) is the energy stored in an object due to its vertical position or height (hh) within a gravitational field.

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The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another. In an isolated system, the total mechanical energy remains constant: Etotal=Ek+EpE_{total} = E_k + E_p.

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The acceleration due to gravity on Earth is represented by gg, which is approximately 9.8 m/s29.8 \, m/s^2 (often rounded to 10 m/s210 \, m/s^2 in some MYP contexts).

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Mechanical Work (WW) is done when a force moves an object, resulting in a change in energy: W=ΔEW = \Delta E.

📐Formulae

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

Ep=mghE_p = mgh

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

v=2Ekmv = \sqrt{\frac{2E_k}{m}}

h=Epmgh = \frac{E_p}{mg}

💡Examples

Problem 1:

A car with a mass of 1200 kg1200 \, kg is traveling at a constant velocity of 20 m/s20 \, m/s. Calculate its kinetic energy.

Solution:

Ek=12×1200 kg×(20 m/s)2=0.5×1200×400=240,000 JE_k = \frac{1}{2} \times 1200 \, kg \times (20 \, m/s)^2 = 0.5 \times 1200 \times 400 = 240,000 \, J (or 240 kJ240 \, kJ)

Explanation:

Substitute the mass (m=1200 kgm = 1200 \, kg) and velocity (v=20 m/sv = 20 \, m/s) into the formula Ek=12mv2E_k = \frac{1}{2}mv^2. Ensure the velocity is squared before multiplying.

Problem 2:

A climber of mass 70 kg70 \, kg ascends a cliff to a height of 50 m50 \, m. Calculate the gain in gravitational potential energy. (Use g=9.8 m/s2g = 9.8 \, m/s^2)

Solution:

Ep=70 kg×9.8 m/s2×50 m=34,300 JE_p = 70 \, kg \times 9.8 \, m/s^2 \times 50 \, m = 34,300 \, J (or 34.3 kJ34.3 \, kJ)

Explanation:

The change in potential energy is calculated using Ep=mghE_p = mgh, where mm is mass, gg is gravity, and hh is the vertical height gained.

Problem 3:

An object is dropped from a height of 20 m20 \, m. Ignoring air resistance, calculate its velocity just before it hits the ground. (Use g=10 m/s2g = 10 \, m/s^2)

Solution:

mgh=12mv2⇒gh=12v2⇒v=2gh=2×10 m/s2×20 m=400=20 m/smgh = \frac{1}{2}mv^2 \Rightarrow gh = \frac{1}{2}v^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 10 \, m/s^2 \times 20 \, m} = \sqrt{400} = 20 \, m/s

Explanation:

Based on the Conservation of Energy, the initial EpE_p at the top is converted entirely into EkE_k at the bottom. The mass mm cancels out from both sides of the equation.