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Energy - Conservation of Energy

Grade 9IB

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

🔑Concepts

The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another or transferred between objects. The total energy in a closed system remains constant (Etotal=constantE_{total} = \text{constant}).

Gravitational Potential Energy (EpE_p) is the energy stored in an object due to its height above a reference point in a gravitational field: Ep=mghE_p = mgh.

Kinetic Energy (EkE_k) is the energy possessed by an object due to its motion: Ek=12mv2E_k = \frac{1}{2}mv^2.

Mechanical Energy (EmechE_{mech}) is the sum of kinetic and potential energy in a system: Emech=Ek+EpE_{mech} = E_k + E_p. In the absence of external forces like friction, ΔEk+ΔEp=0\Delta E_k + \Delta E_p = 0.

Energy is often dissipated as thermal energy (QQ) due to friction or air resistance, which is frequently described as 'lost' energy, though it still exists in the environment.

Efficiency is a measure of how much of the total input energy is converted into useful output energy: Efficiency=Useful energyTotal energy\text{Efficiency} = \frac{\text{Useful energy}}{\text{Total energy}}.

Work (WW) is the transfer of energy that occurs when a force (FF) is applied over a displacement (dd): W=FdW = Fd.

📐Formulae

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

Ep=mghE_p = mgh

W=FΔdW = F \Delta d

P=Wt=ΔEtP = \frac{W}{t} = \frac{\Delta E}{t}

Efficiency=(Useful Energy OutputTotal Energy Input)×100%\text{Efficiency} = \left( \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \right) \times 100\%

Einitial=EfinalE_{initial} = E_{final}

💡Examples

Problem 1:

A 0.5 kg0.5\text{ kg} ball is dropped from a height of 10 m10\text{ m}. Calculate its velocity just before it hits the ground. (Assume g=9.8 m/s2g = 9.8\text{ m/s}^2 and ignore air resistance).

Solution:

Using the conservation of energy: Ep,top=Ek,bottomE_{p, top} = E_{k, bottom}. \n mgh=12mv2mgh = \frac{1}{2}mv^2 \n gimesh=12v2g imes h = \frac{1}{2}v^2 \n 9.8×10=12v29.8 \times 10 = \frac{1}{2}v^2 \n 98=12v298 = \frac{1}{2}v^2 \n v2=196v^2 = 196 \n v=196=14 m/sv = \sqrt{196} = 14\text{ m/s}.

Explanation:

At the top, the ball has maximum gravitational potential energy and zero kinetic energy. As it falls, EpE_p is converted into EkE_k. Just before impact, all EpE_p has become EkE_k.

Problem 2:

An electric motor uses 200 J200\text{ J} of electrical energy to lift a 5 kg5\text{ kg} mass to a height of 3 m3\text{ m}. Calculate the efficiency of the motor. (Use g=9.8 m/s2g = 9.8\text{ m/s}^2)

Solution:

  1. Calculate useful energy output (Work done against gravity): Eout=mgh=5×9.8×3=147 JE_{out} = mgh = 5 \times 9.8 \times 3 = 147\text{ J}. \n 2. Total input energy: Ein=200 JE_{in} = 200\text{ J}. \n 3. Efficiency=147200×100%=73.5%\text{Efficiency} = \frac{147}{200} \times 100\% = 73.5\%.

Explanation:

The efficiency is the ratio of the energy that actually went into lifting the mass (EpE_p) compared to the total energy supplied by the motor. The remaining 26.5%26.5\% was likely dissipated as heat and sound.