R-AEPHY-7

Energy - ability to do work Energy unit: Joules (SI) Work unit: Joules (SI)

It follows that $work = \Delta E$

Potential energy: the potential to do work, but not actually doing work right now.

It also follows $work=F_{s}$.

There are many types of energy.

Law of conservation of energy (same as most other laws of conservation)

• Energy is neither created nor destroyed
• Energy can be transferred = type of energy is unchanged
• Energy can be transformed = type of energy is changed

In motion, we are interested in mechanical energy, i.e. kinetic and gravitational potential.

• $E_{k}$ Kinetic
• $E_{p}$ Gravitational Potential

Kinetic Energy formula: $\frac{1}{2}mv^2$

• $v$ is a vector, and also multiplying a vector by a vector yields a scalar
• It follows kinetic energy is a scalar :)
• $kg \ m^{2} \ s^{-2}$

Gravitational Potential Energy formula:

• $E_{p} = mgh$
• It is scalar
• $kg \ m^{2} / s^{2}$

Applied Force formula:

• $F_{s}=N m$

For objects in motion it is assumed that its total mechanical energy(TME) = Ek + Ep will be constant for that object. It can transfer between the two energies within the object. It is constant for the entire period of motion.

Practice Question: If a man of 15.4 kg is in level flight with a speed of 3.44m/s at a height of 1.58meters above the ground, what is the kinetic, gravitational potential and total mechanical energy

Just sub in the formulas:

• $E_{k}=\frac{1}{2}mv^2=\frac{1}{2}(15.4)(3.44)^2$
• $E_{p}=mgh=(15.4)(9.8)(1.58)$
• $E_{TME}=E_{k}+E_{p}$

"Slightly... well a lot more interesting" Question:

A bird realises that it is a flightless bird and plummets to the ground. What is its final velocity as it hits the ground? It has a mass of $15.4kg$ and falls from a height of $1.58m$. It was "flying" at a speed of $3.44 ms^{-1}$

• Construct vector diagram.

• We know energy is conserved, hence:

  • $E_{k1}+E_{p1}=E_{k2}+E_{p2}$
  • There is no initial kinetic energy.
  • Likewise, when the bird hits the floor it will have no gravitational potential energy.
  • Hence, $E_{p1}=E_{k2}$
  • Hence, $mgh=\frac{1}{2}mv^2$
  • $v=\sqrt{2gh}$
  • $5.56ms^{-1}$

• Looking back at the vector diagram, we know this is the horizontal component. Hence the resultant vector:

  • $\sqrt{ 5.56^{2}+3.44^{2}}=6.54ms^{-1}$
    • This is the final velocity!

• Use trig to find the angle from either the vertical or horizontal.

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